parallelogram law experiment class 11

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Physics practicals class xi, the objective of the experiment :, our objective is to find the weight of a given body using the parallelogram law of vectors..

What does the Parallelogram Law of Vectors state?

If  two vectors acting simultaneously on a particle are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is completely represented in magnitude and direction by the diagonal of that parallelogram drawn from that point.

Parallelogram Law of Vectors explained

On a Gravesand's apparatus, if the body of unknown weight (say S) is suspended from the middle hanger and balancing weights P and Q are suspended from othe two hangers then,

The unknown weight can be calculated from the equation (1).

On a Gravesand's apparatus, if the body of unknown weight (say S) is suspended from the middle hanger and balancing weights

P and Q are suspended from the other two hangers then,

Now construct a parallelogram OACB by assuming a scale (say 1cm=50 gwt) corresponding to the weights P and Q. The diagonal of the parallelogram OC will give the resultant vector. The weight of the unknown body,

If W is the actual weight of the body, then the percentage error in the experiment can be calculated using the equation,

Learning Outcomes

• Students learn what is parallelogram law of vectors.
• They become familiar with the Gravesands apparatus.
• Students are able to find the unknown weight of an object using the parallelogram law of vectors.

Materials Required :

• Parallelogram Law of Forces apparatus (Gravesand's apparatus)
• Two hangers with slotted weights
• A body (a wooden block) whose weight is to be determined
• Thin strong thread
• White drawing paper sheet
• Drawing pins
• Mirror strip
• Sharp pencil
• Half meter scale
• Set squares

Lab Procedure :

• Set up the Gravesand's apparatus and ensure its board is vertical.  This can be tested using the plumb line. Test if the pulleys (let us name them - P1 and Q1) are frictionless. If you feel any friction, oil them.
• Fix the white drawing paper sheet to the board using the drawing pins.
• Take three pieces of strong threads and tie one end of all three together to make a knot. (Let us name this knotted end - O). This knot becomes the junction of the three threads.
• From the other ends of the two threads, tie a weight hanger with the same slotted weights in each; we will name these weights as P and Q.
• From the end of third thread tie the given body, which is the wooden block, which we will name as S.
• Pass the threads with weights P and Q over the pulleys and let the third thread with the block S, stay vertical in the middle of the board.
• The weights P, Q and the wooden block S acts as the three forces along the three threads. At the junction O, the forces are in equilibrium.
• Now adjust the weights P and Q (forces) such that the junction O stays in equilibrium slightly below the middle of the paper.
• See that all the weights hang freely and that none of them touch the board or the table.
• Mark the position of junction O on the paper using a sharp pencil.
• Slightly disturb the weights P and Q and then leave them.
• Once settled, note the position of junction O.  Make sure that this point is very close to the earlier position.
•  Take the mirror strip and keeping it lengthwise under each thread, mark the position of the ends of the image of the thread in the mirror, covering the image by the thread. These new positions are P1, P2 for the thread with the weight P, and Q1 and Q2 for the thread with the weight Q and S1, S2 for the thread with the weight S.
• Remove the paper from the board and with the help of the half metre scale draw lines through the points P1 and P2 to represent P, through points Q1 and Q2 to represent Q and through points S1 and S2 to represent S. These lines must meet at point O.
• Assuming a scale of 1cm = 50 g, mark OA = 3 cm and OB =3 cm to represent P=150g and Q= 150g.
• Complete parallelogram OACB using the set squares and join OC. This represents the resultant vector R which corresponds to the weight S.
• Measure OC and multiply it by the scale (50 g) to get the value of the unknown weight (S).
• For different sets of observation, change P and Q suitably.
• We can find the weight of the wooden block (R) using the equation (1).
• Take the mean of the two values to get the actual weight of the body.
• To find the percentage error in the experiment, measure the actual weight of the body using a spring balance.
• Calculate the percentage error using equation (3).

Observations 

To find the actual weight of the unknown mass, w.

Least count of spring balance = _________g

Zero error of spring balance = ________g

Weight of unknown body by spring balance = ________g

To find the weight of the unknown mass using parallelogram law of vectors

Scale . Let  50 g=1 cm

Calculation

Mean value of unknown weight S = ---------- gwt.

Mean value of unknown weight, R =---------gwt

Unknown weight = (S+R)/2 = ------------gwt= ---------------kgwt

Percentage error = ---------

The unknown weight of given body = ------------------ kgwt .

The result shows the error is within limits of the experiment error.

Precautions

• 1. The board should be stable and ver
• 2. The pulleys should be friction
• 3. The hangers should not touch the board or
• 4. Junction O should be in the middle of the paper
• 5. Points should be marked only when weights are at r
• 6. Points should be marked with sharp
• 7. Arrows should be marked to show direction of for
• 8. A proper scale should be taken to make fairly big parallelogr

Sources of error :

• 1. Pulleys may have
• 2. Weights may not be accurate
• 3. Points may not be marked correctly.
• 4. Weight measured by spring balance may not be much accurate

Viva-Voce [ Parallelogram Law of Vectors ]

Q.1: What are scalar and vector quantities?

Ans. (a) Scalar is a physical quantity which is completely represented only by magnitude with a suitable unit, having no direction, e.g. time, mass, speed, density, work, energy, etc.

(b) Vector is a physical quantity has both ‘magnitude and a specified direction’ e.g. displacement, velocity, acceleration, force, weight, torque, momentum, magnetic field intensity, electric field intensity.

  Q.2: Define resolution of vectors?

Ans. The splitting up of a single vector into two or more vectors is called resolution of vector.

  Q.3: What do you mean by components of a vector?

Ans. Two or more such vectors which are at right angle to each other are called rectangular components.

  Q.4: What are rectangular component?

Ans. The components of a vector which are at right angle to each other are called rectangular components.

  Q.5: Define the following: (a) unit vector (b) null vector (c) position vector (d) negative vector

Ans. (a) Unit vector is that vector whose magnitude is unity ‘1’, and it simply indicates the direction.

(b) Null vector is that vector whose magnitude is zero and it may have any arbitrary direction or no direction. It is also called zero vector.

(c) Position vector is that vector which specifies the position of a point with respect to origin of reference axes.

(d) The negative of a vector is that vector which is equal in magnitude but opposite in direction of that vector.

  Q.6: State head to tail rule of addition of vectors?

Ans. It states that ‘When the representative lines of all the given are drawn, arrange them in such a way that head of first vector line joins with the tail of second vector, then head of second vector joins with the tail of third vector and so on. The line joining the tail of first vector with head of last vector will represent the resultant vector in magnitude and direction’.

  Q.7: State parallelogram law of vector addition?

Ans. It states that ‘If two vectors are completely represented by two adjacent sides of a parallelogram, then the diagonal of the parallelogram from the tails of two vectors gives their resultant vector’.

Q.8: What is a scalar product?

Ans. When the product of two vectors is a scalar quantity it is called scalar product or dot product, e.g. work is a dot product of force and displacement.

Q.9: What is a vector product?

Ans. When the product of two vectors is a vector quantity it is called vector product or cross product, e.g. torque is a vector product of force and force arm.

Q.10: How a vector is multiply by a number?

Ans. When a vector is multiplied by a positive number its magnitude changes but direction remains the same. But when a vector is multiplied by a negative number not only its magnitude changes but its direction is also reversed.

  Question. 11. Can the law of vectors be used to add forces and velocities ?

Answer. Yes, they can be used, because forces and velocities are also vectors.

Question. 12. Why is addition  of vectors different from addition  of scalars ?

Answer. Because vectors have direction also, which makes all the difference

Question. 13. Why is addition  of vectors called composition of vectors ?

Answer. Composition means collection. By addition we collect (convert) many vectors into one single vector.

Question. 14. What is meant by resolution of vectors ?

Answer. It is reverse of composition of vectors. The breaking of a single vector into its components is called resolution of vectors.

Question. 15. What are rectangular components ?

Answer. The two components of a vector, which are perpendicular to each  other, are called rectangular (or right angular)  components.

  Question. 16. What are the main sources of error in the experiment using Gravesand’s apparatus ?

Answer. Its sources of error are

(1) Friction in the pulleys. (2) Weights in the threads.

Question. 17. Why the thread junction  does not come at rest  at same position always  ?

Answer. It is due to friction on the pulleys.

Question. 18. How can this friction be reduced ?

Answer. It is done by oiling the pulleys.

Question. 19. Why the suspended weights are kept free from board  or table ?

Answer. So that their effective weight may not become different due to reaction of board or table.

Question 20.: What is the unit of force?

Ans. The force is measured in Newton (MKS system) or in dyne (CGS system) or in pound (BE system).

Physics is one of the most important subjects in Class 12. As the CBSE exam approaches, students get busy preparing for different subjects. But an essential part of the  CBSE exam  is the practical exams which consist of 30 marks.

Students must know all the experiments along with theorems, laws, and numerical to understand all the concepts of 12th standard physics in a detailed way. Two experiments (8 + 8 marks) are asked from each section in the practical exam. The experiment records and activities consist of 6 marks, the project has 3 marks and viva on the experiment consist of 5 marks.

Science Practicals 11 & 12

Search this blog, class 11 physics practical reading to find the weight of a given body using parallelogram law of vectors., apparatus required.

Observations

Precautions

Sources of error.

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Measurement of the weight of a given body (a wooden block) using the parallelogram law of vector addition

Measurement of the weight of a given body (a wooden block) using the parallelogram law of vector addition.

Apparatus and material required

The given body with hook, the parallelogram law of vector apparatus (Gravesand's apparatus), strong thread, slotted weights (two sets), white paper, thin mirror strip, sharp pencil.

Description of Material

Gravesand's apparatus: It consists of a wooden board fixed vertically on two wooden pillars as shown in Fig. E 5.1 (a). Two pulleys P1 and P2 are provided on its two sides near the upper edge of the board. A thread carrying hangers for addition of slotted weights is made to pass over the pulleys so that two forces P and Q can be applied by adding weights in the hangers. By suspending the given object, whose weight is to be determined, in the middle of the thread, a third force X is applied

Working of this apparatus is based on the parallelogram law of vector addition. The law states that "when two forces act simultaneously at a point and are represented in magnitude and direction by the two adjacent sides of a parallelogram, then the resultant of forces can be represented both in magnitude and direction by the diagonal of the parallelogram passing through the point of application of the two forces. Let P and Q be the magnitudes of the two forces and θ the angle between them. Then the resultant R of P and Q is given by

R = √ P 2 + Q 2 + 2PQ cos θ

Weight of body O in one pan = Weight of body O' in other pan

Or, mg = m s g

where g is the acceleration due to gravity, which is constant. Thus,

the mass of object O in one pan = standard mass in the other pan

• Set the board of Gravesand's apparatus in vertical position by using a plumb-line. Ensure that the pulleys are moving smoothly. Fix a sheet of white paper on the wooden board with drawing pins.
• Take a sufficiently long piece of string and tie the two hangers at its ends. Tie another shorter string in the middle of the first string to make a knot at 'O'. Tie the body of unknown weight at the other end of the string. Arrange them on the pulley as shown in Fig. E 5.1 (a) with slotted weights on the hangers.
• Add weights in the hangers such that the junction of the threads is in equilibrium in the lower half of the paper. Make sure that neither the weights nor the threads touch the board or the table.
• Bring the knot of the three threads to position of no-friction. For this, first bring the knot to a point rather wide off its position of no-friction. On leaving there, it moves towards the position of no-friction because it is not in equilibrium. While it so moves, tap the board gently. The point where the knot thus come to rest is taken as the position of no-friction, mark this point. Repeat the procedure several times. Each time let the knot approach the position of no-friction from a different direction and mark the point where it comes to rest. Find by judgement the centre of those points which are close together. Mark this centre as O.
• To mark the direction of the force acting along a string, place a mirror strip below the string on the paper . Adjust the position of the eye such that there is no parallax between the string and its image. Mark the two points A1 and A2 at the edges of the mirror where the image of the string leaves the mirror [Fig E 5.1 (b)]. Similarly, mark the directions of other two forces by points B1 and B2 and by points X1 and X2 along the strings OB and OX respectively.
• Remove the hangers and note the weight of each hanger and slotted weights on them.
• Place the board flat on the table with paper on it. Join the three pairs of points marked on the paper and extend these lines to meet at O. These three lines represent the directions of the three forces.
• Choose a suitable scale, say 0.5 N (50 g wt) = 1cm and cut off length OA and OB to represent forces P and Q respectively acting at point O. With OA and OB as adjacent sides, complete the parallelogram OACB. Ensure that the scale chosen is such that the parallelogram covers the maximum area of the sheet.
• Join points O and C. The length of OC will measure the weight of the given body. See whether OC is along the straight line XO. If not, let it meet BC at some point C'. Measure the angle COC'.
• Repeat the steps 1 to 9 by suspending two different sets of weights and calculate the mean value of the unknown weight.

Observations

Weight of each hanger = ... N

Scale, 1cm = ... N

Table E 5.1: Measurement of weight of given body

The weight of the given body is found to be ... N.

Precautions

• Board of Gravesand's apparatus is perpendicular to table on which it is placed, by its construction. Check up by plumb line that it is vertical. If it is not, make table top horizontal by putting packing below appropriate legs of table.
• Take care that pulleys are free to rotate, i.e., have little friction between pulley and its axle.

Sources of Error

• Friction at the pulleys may persist even after oiling.
• Slotted weights may not be accurate.
• Slight inaccuracy may creep in while marking the position of thread.

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Class 11 Physics Lab Experiment list

• 1 Force of Limiting Friction vs Normal Reaction Experiment
• 2 Simple Pendulum experiment to plot L-T graphs
• 3 Measurement of the weight of a given body (a wooden block) using the parallelogram law of vector addition
• 4 To determine mass of two different objects using a beam balance
• 5 To determine the radius of curvature of a given spherical surface by a spherometer
• 6 Screw gauge experiment to measure diameter of wire
• 7 Vernier callipers experiment to measure diameter of spherical or cylindrical body

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• Electrical and Electronics
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CBSE Class 11 Lab Manual for Chapter 5 2 To Find the Weight of a Given Body Using Parallelogram Law of Vectors PDF Download

CBSE Class 11 Lab Manual Chapter 5 2 To Find the Weight of a Given Body Using Parallelogram Law of Vectors Download here in pdf format. These Lab Manual may be freely downloadable and used as a reference book. Learning does not mean only gaining knowledge about facts and principles rather it is a path which is informed by scientific truths, verified experimentally. Keeping these facts in mind, CBSE Class 11 Lab Manual for Chapter 5 2 To Find the Weight of a Given Body Using Parallelogram Law of Vectors have been planned, evaluated under subject Improvement Activities. Check our CBSE Class 11 Lab Manual for Chapter 5 2 To Find the Weight of a Given Body Using Parallelogram Law of Vectors. We are grateful to the teachers for their constant support provided in the preparation of this CBSE Class 11 Lab Manual.

CBSE Class 11 Lab Manual for Chapter 5 2 To Find the Weight of a Given Body Using Parallelogram Law of Vectors

The laboratory is important for making the study complete, especially for a subject like Science and Maths. CBSE has included the practicals in secondary class intending to make students familiarised with the basic tools and techniques used in the labs. With the help of this, they can successfully perform the experiments listed in the CBSE Class 11 Lab Manual.

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By performing the experiments, students will know the concept in a better way as they can now view the changes happening in front of their eyes. Their basics will become solid as they will learn by doing things. By doing this activity they will also get generated their interest in the subject. Students will develop questioning skills and start studying from a scientific perspective. Here we have given all the necessary details that a Chapter 5 2 To Find the Weight of a Given Body Using Parallelogram Law of Vectors student should know about CBSE Class 11 Lab Manual. From CBSE Science practical to Lab manual, project work, important questions and CBSE lab kit manual, all the information is given in the elaborated form further in this page for Chapter 5 2 To Find the Weight of a Given Body Using Parallelogram Law of Vectors students.

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• Parallelogram Law of Vectors

To Find the Weight of a Given Body Using Parallelogram Law of Vectors

This law can be explained as, “If two forces acting simultaneously on a particle are represented in magnitude and direction by the two adjacent sides of the parallelogram, the diagonal of that parallelogram will be expressed as the resultant of these two forces represented in direction and magnitude.”

(Image will be Uploaded Soon)

The main objective of this experiment is to find the weight of the given object (body) applying the law of forces.

This is the basic law followed by basic mechanics. Its applications are used as lifting loads of cranes and bracket stay wires, etc.

To conduct this experiment, we need some essential apparatus such as;

Gravesand's apparatus which is an ideal apparatus for parallelogram law of forces

An object with an unknown weight (used for identifying its weight)

Slotted weights are hung with two hangers

The thread which is thin as well as durable

White colour drawing sheet

Pins to hook up drawing sheet

Pointed pencil (2HB)

Mirror strip

Set squares

Half-meter scale

To Find the Weight of a Given Body Using Parallelogram

It can be calculated by the use of Gravesand's apparatus. The concept is that the vector sum of the forces experienced by the two masses hanging on the pulley is equal to the force of the object hanging in the middle. The same force is experienced by the mass in the middle.

If an anonymous weight body (S) is suspended from the centre of the hanger, and P and Q are the two symmetric weights from the other end of the hanger, then that unknown weight can be calculated by using the equation below;

\[S=\sqrt{P^2+Q^2+2PQcos\theta }\]

\[\vec{P},\vec{Q}\] are two identical forces

The unknown weight can be termed S

P and Q are the balance weights used in the experiment

θ is the angle between two forces

To Find the Weight of a Given Body Using Parallelogram Law of Vector. We need to follow certain steps to do so:

Gravesand's apparatus is set up with a board vertically with the help of a plumb line.

P 1 and P 2 pulleys should be oiled properly to make them frictionless

Fix the white sheet on the board with the help of drawing pins.

“O” is the knot shaped 

P and Q are the weights that are tied up at both the ends of the hanger and S be the third body tied at the third end.

Junction O should be sustained at equilibrium by maintaining weights P and Q.

P, Q, and S these three weights act as three forces \[\vec{P},\vec{Q}\]and\[\vec{S}\]

These weights should be hung freely without making any contact with the board.

Mark the position of the junction of O with the help of a dark pencil.

Disturb the weights at P and Q and leave them free.

The position of junction O will be closed as compared to the earlier position.

Let the position of P be P 1 and P 2 , Q 1 and Q 2 will be the position of Q and S 1 and S 2 will be the position of S. All these positions are being written down with the use of a mirror.

By taking a scale, 1cm =50gm

OA = 3cm and

These parameters are taken to represent 

P = 150 gm and Q = 150 gm

Where R is represented by finishing the parallelogram OACB and by drawing OC line with the use of set squares

When measuring OC, the result shows 3.9cm.

P and Q can be altered for different sets.

By utilizing spring balance, calculate the weight of the wooden box.

To Find the Weight of a Given Body Using Parallelogram Law of Vectors Observation Table

Least count of spring balance = …… g

Zero error of spring balance = …….. G

Weight of unknown body by spring balance = …….g

Scale used: Let 1 cm = 50 g

Calculations

After all the measurements

OC = 3.9cm, R = 50 * 3.9 = 195 g

Unknown weight is calculated as, S =195 g

Mean unknown weight will be S = \[\frac{(S_1+S_2+S_3)}{3}\] = 195 g

Weight measured from spring balance = 200 g

So the error is calculated through the difference between weight measured and mean unknown weight such as;

200g-195g = 5g

The error in this experiment is under its limits as per experimental error.

The unknown weight of the given body = 195g

The error is within the limits of experimental error.

Learn about Precautions

The board used for the experiment should be placed vertically and stable.

Try to make these pulleys friction-free.

The table and board should not make any contact with the hangers.

The junction O should lie in the middle of the paper

The points should be marked when weights are stationary.

A sharp pencil (2HB) should be useful to mark all the points.

Arrows should be indicated to show the direction of forces.

A proper scale should be used for making a fairly big parallelogram.

To Evaluate Sources of Error

Friction in the pulleys might cause an error.

The accuracy of weights might vary.

The marked point may be correct.

The accuracy of weight obtained from spring balance may not be accurate.

Learning Outcomes

Students learn exactly what the parallelogram law of vectors is.

Gravesand's apparatus is familiarized with them.

Using the parallelogram law of vectors, students can find the unknown weight of an object.

FAQs on Parallelogram Law of Vectors

1. What is the Definition of the Parallelogram Law of Vector Addition?

If two vectors act simultaneously to represent the magnitude and direction as the two sides of a parallelogram, then the diagonal is depicted as the resultant of these two vectors. 

2. What is the Rule of the Parallelogram?

The easiest form of parallelogram law (also called parallelogram identity) belongs to geometry in mathematics. It states that the summation of squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals.

3. What are the Examples of the Parallelogram?

A parallelogram consists of four sides, and these sides opposite each other are parallel, i.e. they don’t intersect. Some examples are squares, rhombuses, and rectangles.

4. How to Calculate the Mean Unknown Weight in Gravesand’s Apparatus?

Suppose we are getting three unknown weights such as:

S 1 = 195g; S 2 = 195.5g; S 3 = 194.5g 

The unknown mean weight will be, 

S = (S 1 +S 2 +S 3 )/3 = (195+195.5+194.5)/3 = 195g

5. Define the addition of vectors?

The operation of adding two or more vectors together to form a vector sum is known as the addition of vectors.

6. Why is the addition of vectors different from the addition of scalars?

The addition of a vector has both direction and magnitude, whereas adding a scalar only has magnitude. As a result, adding a vector is not the same as adding a scalar.

7. How a vector is multiplied by a number?

The magnitude of a vector changes when it is multiplied by a positive number, but the direction remains the same. When a vector is multiplied by a negative number, not only does its magnitude change but also its direction is reversed.

8. Define the following: (a) unit vector (b) null vector (c) position vector (d) negative vector

The magnitude of a unit vector is unity '1', and it just denotes the direction.

A null vector is one whose magnitude is zero and which can have any arbitrary direction or none at all. It's also known as a zero vector.

A position vector is a vector that specifies a point's position concerning the origin of reference axes.

The negative of a vector is the vector that is equal in magnitude but moves in the opposite direction of that vector.

9.What are scalar and vector quantities?

A scalar is a physical quantity that has no direction and is completely represented only by magnitude with a suitable unit, e.g. time, mass, speed, density, work, energy, and so on.

A vector is a physical quantity that has both "magnitude and a specified direction," for example, displacement, velocity, acceleration, force, weight, torque, momentum, magnetic field intensity, and electric field intensity.

10.What are the main sources of error in the experiment using Gravesand’s apparatus?

 Its sources of error are:

Friction in the pulleys.

Weights in the threads

11.What is the importance of practicals?

One of the most important subjects is physics. As the CBSE exam approaches, students become busy studying various subjects. However, the practical exams, which are of 30 marks, are a significant part of the CBSE exam.

To understand all the topics of 12th-grade physics in detail, students must know all the experiments as well as theorems, laws, and numerical. In the practical exam, two experiments (8 + 8 marks) are required from each section. Six marks are awarded for experiment records and activities, three marks are given for the project, and viva on the experiment consists of 5 marks.

The Physics practicals For Class 12 CBSE are provided here so that students can better understand the experiments. Before experimenting, students should study the theory and law behind it. Also, go through the viva voce questions and answers for each experiment that are available on the website.

12.Does Vedantu provide NCERT solutions?

Vedantu is a platform that provides NCERT Solutions and other study materials for students for free. Students who are looking for maths solutions can download the Class 10 Maths NCERT Solutions to help you to revise the complete syllabus and score more marks in your examinations. Science Students who are also looking for NCERT Solutions for Class 10 Science, can find the Solutions curated by our expert teachers. They are really helpful.

13.Are the previous year papers helpful?

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We hope you found this article helpful. For more such articles, visit the site of Vedantu. If you have any doubts related to this, let us know in the comments section below. We will be happy to help you.

Parallelogram Triangle Law of vector addition Experiment Class 11 | Gravesand’s apparatus

To find the weight of a given body (Wooden Block) using parallelogram law of vectors.

• Parallelogram Apparatus (Gravesand’s Apparatus)
• Two Slotted Weights  With Hanger
• Wooden Block With Hook
• Spring Balance
• Mirror Strip
• Cotton Thread (roll)
• Drawing Pin

In Fig. 5.1 we see the Gravesand’s apparatus or Parallelogram apparatus.  It consists of a wooden board A fixed vertically on two pillars. There are two pulleys P and Q fitted at the same level at the top of the board. Three set of slotted weights are supplied with the apparatus which can be used to verify the parallelogram law of vectors.  A thread carrying hangers for addition of slotted weights is made to pass over the pulleys so that two forces P and Q can be applied by adding weights in the hangers. By suspending the given object, whose weight is to be determined, in the middle of the thread, a third force W is applied.

If  two vectors acting simultaneously on a particle are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is completely represented in magnitude and direction by the diagonal of that parallelogram drawn from that point.

Let two vectors  (P) ⃗  and (Q) ⃗  act simultaneously on a particle O at an angle θ as shown in figure 5.2. They are represented in magnitude and direction by the adjacent sides OA and OB of a parallelogram OACB drawn from a point O. Then the diagonal OC, will represent the resultant R in magnitude and direction. Mathematically we can say, 

(R) ⃗= (P) ⃗ + (Q) ⃗

If a body of unknown weight W is suspended from the middle hanger and balancing weights P and Q are suspended from other two hangers then, (R) ⃗ and the three vectors (P) ⃗,(Q) ⃗ ,(W) ⃗  are in equilibrium. Under equilibrium  |W| ⃗= |R| ⃗. Weight of a body is a force.

Hence, |W| ⃗  = |P| ⃗+|Q| ⃗.

If  S  is the actual weight of the body, then the percentage error in the experiment can be calculated using 

Procedure :

• Set the Gravesand’s apparatus with its board in vertical position with the help of plumb line.
• Ensure that the pulleys are moving smoothly. Oil them if necessary.
• Fix a drawing paper sheet on the board with the help of drawing pins.
• Take a long thread and tie two hangers H1 and H2 to the both ends of the thread.
• Tie the given body B, whose weight is to be found, with one end of another shorter thread.
• Tie the other end of the shorter thread in the middle of the longer thread to form a small knot, O. This knot becomes the junction of the three threads.
• Pass the longer threads over the two pulley P1 and P2 of the apparatus and place two equal slotted weights P and Q the hangers H1 and H2 .
• The body B is made to hang vertically with the knot O somewhere in the drawing paper.
• Adjust the weights P and Q to keep the knot O at a position slightly below the middle of the paper. This is the equilibrium position of O. Ensure that all the weights hang freely and do not touch the board.
• Mark the position of junction O on the white paper by a pencil.
• Ensure that there is no friction at the two pulleys P1 and P2. For that purpose, disturb the positions of P and Q a little and leave them. The knot should go back to its initial equilibrium position, If it does not, oil the pulleys to remove friction.
• Mark the position of the equilibrium of the knot (O) by a dot of a pencil on the paper sheet.
• Take the thin mirror strip m and place it breadthwise on the paper under one of the threads.  See the image of the thread in the mirror. Adjust your line of sight such that it is at right angle to strip and the thread. At this point cannot see the image of the thread (meaning, parallax is removed). Mark two points P1 and P2 on the paper, on either side of the strip exactly behind the thread.
• Similarly, mark two points Q1, Q2 and W1, W2 for the other two threads.
• Remove the threads and hangers from the apparatus and note the weights of the two hanger H1 and H2 along with the slotted weights in table 5.1. Let these two weights be P and Q.
• Remove the paper from the board and join P1 and P2, Q1, Q2 are W1, W2 with the help of ruler to get three lines which should meet O. These three lines represent the directions of three forces.
• To represent the vector, chose a suitable scale, say 10 gm (0.1N) = 1 cm.
• Cut the lines OA and OB to represent the forces  and  respectively. For example, if P = 60 gm and Q = 60 gm, then OA = 6 cm and OB = 6 cm.
• With OA and OB as adjacent sides, complete the parallelogram OACB as shown in fig 5.2.
• Join O and C. The length OC represents the resultant vector  which corresponds to the unknown weight W.
• Measure OC ( l ) and multiply it by the scale (10 g) to get the value of R.
• Measure the angle  () using a protractor and calculate W using equation (i).
• Repeat the experiment with fresh paper and using different weights (un-equal) P and Q in the hangers.
• Find the calculated value of weight R.
• Measure the actual weight (S) of the unknown weight (W) by the spring balance and calculate the percentage error using equation (ii).

Observation:

Least count of spring balance, L.C. = _______________ gm

Weight of B by spring balance, S = ________________gm

Scale factor: Let ______ cm = ___________gm

Table 5.1 Measurement of weight of given body

Calculations:

For equal weights

Weight of unknown body by observation,  W:   ……………………….

Weight of unknown body by calculation,  W ‘ : ……………………….

For un-equal weights

Weight of unknown body by observation,  W: ……………………….

Weight of unknown body by observation,  W ‘ : ……………………….

Precautions :

• The pulleys should be frictionless.
• The boards should be stable and vertical.
• The hangers should not touch the board.
• There should be one central knot and it should be small.
• When the lines of action of the forces are marked, the hangers should at rest.
• The scale should be so chosen that a fairly large parallelogram is obtained.
• Threads should be strong and thin, so that they may be regarded as massless.
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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

To Find the Weight of a Given Body Using Parallelogram Law of Vectors

November 23, 2016 by Bhagya

Aim  To find the weight of a given body using parallelogram law of vectors.

Apparatus  Parallelogram law of forces apparatus (Gravesand’s apparatus), plumb line, two hangers with slotted weights, a body (a wooden block) whose weight is to be determined, thin strong or thread, white drawing paper sheet, drawing pins, mirror strip, sharp pencil, half metre scale, set squares, protractor.

• Set up the Gravesand’s apparatus with its board vertical, tested with the help of a plumb line.
• Test that pulleys P 1  and P 2 are frictionless. Oil them if necessary.
• Fix the white drawing paper sheets on the board with the help of drawing pins.
• Take three pieces of strong thread and tie their one end together to make knot O. This knot becomes junction of the three threads.
• From the other ends of two threads, tie a hanger with some slotted weights in each. These serve as the weights P and Q. From the other end of third thread tie the given body S.
• Pass threads with weights P and Q over the pulleys and let the third thread with given body S, stay vertical in the middle of the board.
• Adjust the weights P and Q (forces) such that the junction O stays in equilibrium slightly below the middle of the paper.

• See that all the weights hang freely and none of them touches the board or the table.
• Mark the position of junction O on the white paper sheet by a sharp pencil.
• Disturb weights P and Q and leave them.
• Note position of junction O. It must be very close to earlier position. (If not, oil the pulleys to remove friction.)

• Remove paper from the board.
• With the help of a half metre scale draw lines through points P 1  and P 2 to represent P, through points Q 1  and Q 2 to represent Q and through points S 1  and S 2 to represent S. These lines must meet at point O.
• Taking a scale, 1 cm = 50 g, take OA = 3 cm and OB = 3 cm to represent P = 150 g and Q = 150 g.
• Complete parallelogram OACB using set squares and join OC. It represents R.
• Measure OC. It comes to be 3.9 cm.
• For different sets of observation, change P and Q suitably.
• Find weight of the wooden block by a spring balance.

Result  The unknown weight of given body = 195 g The error is within limits of experiment error.

Precautions 

• The board should be stable and vertical.
• The pulleys should be friction less.
• The hangers should not touch the board or table.
• Junction O should be in the middle of the paper sheet.
• Points should be marked only when weights are at rest.
• Points should be marked with sharp pencil.
• Arrows should be marked to show direction of forces.
• A proper scale should be taken to make fairly big parallelogram.

Sources of error 

• Pulleys may have friction.
• Weights may not be accurate.
• Points may not be marked correctly.
• Weight measured by spring balance may not be much accurate.

Question. 1. Define a scalar quantity. Answer. Read Art. 5.01.

Question. 2. Define a vector quantity. Answer. Read Art. 5.01.

Question. 3. How a geometrical vector represents a vector quantity ? Answer. Read Art. 5.02.

Question. 4. Define addition of vectors. Answer. Read Art. 5.03.

Question. 5. State parallelogram law of addition of two vectors. Answer. Read Art. 5.04 (a).

Question. 6. Write expression for magnitude and direction of the resultant of two vectors. Answer. The required expression are

Question. 7. State triangle law of addition of two vectors. Answer . Read Art. 5.05 (a).

Question. 8. Define equilibrium of vectors. Answer. Read Art. 5.06.

Question. 9. Define an equilibrant vector. Answer. Read Art. 5.06.

Question. 10. State triangle law for equilibrium of three vectors. Answer. Read Art. 5.07 (a).

Question. 11. Can the law of vectors be used to add forces and velocities ? Answer. Yes, they can be used, because forces and velocities are also vectors.

Question. 12. Why is addition of vectors different from addition of scalars ? Answer. Because vectors have direction also, which makes all the difference.

Question. 13. Why is addition of vectors called composition of vectors ? Answer. Composition means collection. By addition we collect (convert) many vectors into one single vector.

Question. 14. What is meant by resolution of vectors ? Answer. It is reverse of composition of vectors. The breaking of a single vector into its components is called resolution of vectors.

Question. 15. What are rectangular components ? Answer. The two components of a vector, which are perpendicular to each other, are called rectangular (or right angular) components.

Question. 16. What are the main sources of error in the experiment using Gravesand’s apparatus ? Answer. Its sources of error are (1) Friction in the pulleys. (2) Weights in the threads.

Question. 17. Why the thread junction does not come at rest at same position always ? Answer. It is due to friction on the pulleys.

Question. 18. How can this friction be reduced ? Answer. It is done by oiling the pulleys.

Question. 19. Why the suspended weights are kept free from board or table ? Answer. So that their effective weight may not become different due to reaction of board or table.

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Parallelogram law of vector addition Questions and Answers

Note: vectors are shown in bold. scalars are shown in normal type

The diagram above shows two vectors A and B with angle p between them.

R is the resultant of A and B

This is the resultant in vector

R is the magnitude of vector R

Similarly A and B are the magnitudes of vectors A and B

R = √(A 2 + B 2 2ABCos p) or [A 2 + B 2 2ABCos p] 1/2

To give the direction of R we find the angle q that R makes with B

Tan q = (A Sin p)/(B + A Cos q)

A vector is completely defined only if both magnitude and direction are given.

Two forces of 3 N and 4 N are acting at a point such that the angle between them is 60 degrees. Find the resultant force

Magnitude R of the resultant force is R = √(3 2 + 4 2 + 2 x 3 x 4 Cos 60 deg)

= √(9 + 16 + 12) = √(37 = 6.08 N

Direction of R is given by finding the angle q

tan q = (3 Sin 60 deg)/(4 + 3 Cos 60 deg) = 0.472

q = tan -1 0.472

Thus R is 6.08 N in magnitude and is at an angle of 25.3 deg to the 4 N force.

A car goes 5 km east 3 km south, 2 km west and 1 km north. Find the resultant displacement.

First we will make the vector diagram

O to A 5 km east

A to B 3 km south

B to C 2 km west

c to D 1 km north

Net displacement is OD

Along the horizontal direction: 5 km east - 2 km west = 3 km east

Along the vertical direction: 3 km south - 1 km north = 2 km south

OD = √(3 2 + 2 2 + 2 x 3 x Cos 90 deg)

= √(3 2 + 2 2 )

tan p = 2/3

or p = tan -1 2/3 = 34 deg

Thus resultant displacement is 3.6 km, 34 deg south of east.

Rectangular components of a vector

To find the component of a vector along a given axis, we drop a perpendicular on the given axis from the vector

For example OA is the given vector. We have to find its component along the the horizontal axis. Let us call it x-axis. We drop a perpendicular AB from A onto the x-axis. The length OB is the component of OA along x-axis. If OA makes angle p with the horizontal axis, then in triangle OAB, OB/OA = Cos P or OB = OA Cos P.

Remember that component of a vector is a scalar quantity. If the component is along the negative direction, we put a (-) sign with it.)

Usually we resolve the vector into components along mutually perpendicular components.

OB is the x component OB = OA Cos p.

Similarly component along the vertical direction or the y axis is OC

OCAB is a rectangle.

look at triangle OAB again,

AB/OA = Sin p

=> AB = OA Sin p = OC

Thus y component OC = OA Sin p.

Note that p is the angle with the horizontal axis.

Find the x and y components of a 25 m displacement at an angle of 210 deg.

OA is the displacement vector. The angle with the horizontal axis is 210 deg - 180 deg = 30 deg

x component = OB = -25 Cos 30 deg = -21.7

y component = AB = -25 Sin 30 deg = -12.5 m

Note that each component is pointing along the negative coordinate direction and thus we must take it as negative.

Now we will solve a problem using the component method

Find the resultant of the following two displacements: 2 m at 30 deg and 4 m at 120 deg. The angles are taken relative to the x axis.

Rx = 2 Cos 30 deg - 4 Cos 60 deg = - 0.268 m

Ry = 2 Sin 30 deg + 4 Sin 60 degg = 4.46 m

R = √(Rx 2 + Ry 2 )

= √(-0.268 2 + 4.46 2 ) = 4.47 m

tan q = Ry/Rx = 4.46/0.268

=> q = 86.6 deg

p = 180 deg - 86.6 deg = 93.4 deg

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Parallelogram Law

In Mathematics, the parallelogram law is the fundamental law that belongs to elementary Geometry. This law is also known as parallelogram identity. In this article, let us look at the definition of a parallelogram law, proof, and parallelogram law of vectors in detail.

Parallelogram Law of Addition

The Parallelogram law states that the sum of the squares of the length of the four sides of a parallelogram is equal to the sum of the squares of the length of the two diagonals. In Euclidean geometry , it is necessary that the parallelogram should have equal opposite sides.

If ABCD is a parallelogram, then AB = DC and AD = BC. Then according to the definition of the parallelogram law, it is stated as

2(AB) 2 + 2 (BC) 2 = (AC) 2 + (BD) 2 .

In case the parallelogram is a rectangle, then the law is stated as:

2(AB) 2 + 2 (BC) 2 = 2(AC) 2

This is because in a rectangle, two diagonals are of equal lengths. i.e., (AC = BD)

Parallelogram Law of Vectors

If two vectors are acting simultaneously at a point, then it can be represented both in magnitude and direction by the adjacent sides drawn from a point. Therefore, the resultant vector is completely represented both in direction and magnitude by the diagonal of the parallelogram passing through the point.

Consider the above figure,

The vector P and vector Q represents the sides, OA and OB, respectively.

According to the parallelogram law, the side OC of the parallelogram represents the resultant vector R.

Parallelogram Law of Addition of Vectors Procedure

The steps for the parallelogram law of addition of vectors are:

• Draw a vector using a suitable scale in the direction of the vector
• Draw the second vector using the same scale from the tail of the first vector
• Treat these vectors as the adjacent sides and complete the parallelogram
• Now, the diagonal represents the resultant vector in both magnitude and direction

Parallelogram Law Proof

Let AD=BC = x, AB = DC = y, and ∠ BAD = α

Using the law of cosines in the triangle BAD, we get

x 2 + y 2 – 2xy cos(α) = BD 2 ——-(1)

We know that in a parallelogram, the adjacent angles are supplementary. So

∠ADC = 180 – α

Now, again use the law of cosines in the triangle ADC

x 2 + y 2 – 2xy cos(180 – α) = AC 2 ——-(2)

Apply trigonometric identity cos(180 – x) = – cos x in (2)

x 2 + y 2 + 2xy cos(α) = AC 2

Now, the sum of the squares of the diagonals (BD 2 + AC 2 ) are represented as,

BD 2 + AC 2   = x 2 + y 2 – 2xycos(α) + x 2 + y 2 + 2xy cos(α)

Simplify the above expression, we get;

BD 2 + AC 2 =2x 2 + 2 y 2 ——-(3)

The above equation is represented as:

BD 2 + AC 2 = 2(AB) 2 + 2(BC) 2

Hence, the parallelogram law is proved.

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What is parallelogram in physics class 11?

The Parallelogram law states that the sum of the squares of the length of the four sides of a parallelogram is equal to the sum of the squares of the length of the two diagonals.

Table of Contents

How do you do the parallelogram law of vectors experiment?

Let P1 and P2 be the position of P, Q1 and Q2 be the position of Q and S1 and S2 be the position of S which are taken down with the help of the mirror. Remove the paper from the board. Using half-meter scale draw lines through P1 and P2, Q1 and Q2 and S1 and S2 represent P, Q, and S respectively.

What is parallelogram method in physics?

The parallelogram rule says that if we place two vectors so they have the same initial point, and then complete the vectors into a parallelogram, then the sum of the vectors is the directed diagonal that starts at the same point as the vectors.

What is the principle of parallelogram?

The principle of the parallelogram of forces states that if you have two forces, the resultant force between these two forces can be calculated by placing the tails of two forces on the same point and constructing a parallelogram by doing a mirror image of these two vectors.

What is parallelogram law of velocity?

Statement of Parallelogram Law of Vector Addition: If two vectors can be represented by the two adjacent sides (both in magnitude and direction) of a parallelogram drawn from a point, then their resultant sum vector is represented completely by the diagonal of the parallelogram drawn from the same point.

How do you verify a triangle law of forces?

This law can also be stated as: If two forces acting on a particle represented in magnitude and direction by the two sides of the triangle taken in order then their resultant will be given by the third side of the triangle taken in opposite direction.

What is triangle law of vector addition?

Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

How do you find the force of a parallelogram?

How do you find the angle of a parallelogram?

How do you find a vector of a parallelogram?

Who discovered parallelogram?

His picks for 2014 are two articles on the life and Parallelogram Theorem of French mathematician Pierre Varignon (1654-1722), including ideas for how the Parallelogram Theorem can be explored and extended in the classroom. Parallelogram EFGH is formed by connecting the midpoints of the sides of quadrilateral ABCD.

What is the vector rule?

Two vectors are equal only if they have the same magnitude and direction. This condition can be described mathematically as follows: Vector is equal to vector only when. 5. When two or more vectors are added together, the resulting vector is called the resultant.

What is resultant of vector?

A resultant vector is defined as a single vector that produces the same effect as is produced by a number of vectors collectively.

What is the vector sum?

A vector sum is the result of adding two or more vectors together via vector addition. It is denoted using the normal plus sign, i.e., the vector sum of vectors , , and is written .

What is the formula of resultant vector?

R = A + B. Formula 2 Vectors in the opposite direction are subtracted from each other to obtain the resultant vector. Here the vector B is opposite in direction to the vector A, and R is the resultant vector.

What are the 4 types of parallelograms?

What are the 4 types of parallelogram? There are 4 types of parallelograms, including 3 special types. The four types are parallelograms, squares, rectangles, and rhombuses.

What are the 7 properties of parallelogram?

• Opposite sides are parallel.
• Opposite sides are congruent.
• Opposite angles are congruent.
• Same-Side interior angles (consecutive angles) are supplementary.
• Each diagonal of a parallelogram separates it into two congruent triangles.
• The diagonals of a parallelogram bisect each other.

What is law of parallelogram of vector addition?

– Parallelogram law of vector addition states that. if two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of the two vectors is given by the vector that is diagonal passing through the point of contact of two vectors.

What is a vector diagram?

Vector diagrams are diagrams that depict the direction and relative magnitude of a vector quantity by a vector arrow. Vector diagrams can be used to describe the velocity of a moving object during its motion. For example, a vector diagram could be used to represent the motion of a car moving down the road.

What is resultant of a force?

The resultant force is described as the total amount of force acting on the object or body along with the direction of the body. The resultant force is zero when the object is at rest or it is traveling with the same velocity as the object.

What is universal force table?

₹ 4,500/ Set(s) Get Latest Price. A force board (or force table) is a common physics lab apparatus that has three (or more) chains or cables attached to a center ring. The chains or cables exert forces upon the center ring in three different directions.

What is the law of polygon?

Polygon law of vector addition states that if a number of vectors can be represented in magnitude and direction by the sides of a polygon taken in the same order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order.

Is parallelogram law can be proved experimentally?

Also, both OC and OD are acting in the opposite direction. ∠COD must be equal to 180o. If OC = OD and ∠COD = 180o, one can say that parallelogram law of force is verified experimentally.

What is a negative vector?

A negative of a vector represents the direction opposite to the reference direction. It means that the magnitude of two vectors are same but they are opposite in direction. For example, if A and B are two vectors that have equal magnitude but opposite in direction, then vector A is negative of vector B.

What is the definition of zero vector?

Definition of zero vector : a vector which is of zero length and all of whose components are zero.

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What is stable and unstable equilibrium.

Stable equilibrium exists when the object is in its lowest energy condition; metastable equilibrium exists when additional energy (ΔG) must be introduced before the object can…

What is traction and friction?

While friction is a general physical expression, vehicle traction can be defined as the friction between a drive wheel and the road surface. “traction is the…

What is positive and negative force?

Note: Force can be positive or negative. If there is negative acceleration, the force will be negative. Usually, Force working toward the right is considered as…