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.”
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Aim
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.
Apparatus
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)
Plumb line
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
Protractor
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.
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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;
\[S=\sqrt{P^2+Q^2+2PQcos\theta }\]
Where,
\[\vec{P},\vec{Q}\] 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
Procedure
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.
P1 and P2 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 P1 and P2, Q1 and Q2 will be the position of Q and S1 and S2 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
OB= 3cm
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.
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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.
Results
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 Parallelogram Law of Vector Addition as per the CBSE Class 11 syllabus?
The Parallelogram Law of Vector Addition states that if two vectors, acting simultaneously at a point, are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, then their resultant vector is represented in both magnitude and direction by the diagonal of the parallelogram passing through the same point.
2. What is the formula to find the magnitude and direction of the resultant vector using this law?
If two vectors A and B have an angle θ between them, the magnitude (R) and direction (α) of their resultant vector R are given by the following formulas:
- Magnitude (R): R = √[A² + B² + 2AB cos(θ)]
- Direction (α): tan(α) = [B sin(θ)] / [A + B cos(θ)], where α is the angle the resultant vector R makes with vector A.
3. How does the Parallelogram Law of Vectors differ from the Triangle Law of Vector Addition?
Both laws are used to find the resultant of two vectors, but they differ in their graphical representation:
- Arrangement: In the Triangle Law, vectors are arranged head-to-tail. The second vector starts from the head of the first vector. In the Parallelogram Law, vectors are arranged tail-to-tail, meaning they are co-initial (start from the same point).
- Resultant: In the Triangle Law, the resultant is the third side of the triangle, closing the loop. In the Parallelogram Law, the resultant is the diagonal of the parallelogram that starts from the common initial point.
Ultimately, both methods yield the exact same resultant vector.
4. Can you provide a real-world example of the Parallelogram Law in action?
A classic example is a boat crossing a river. The boat has a velocity vector pointing across the river, and the river current has a velocity vector pointing downstream. These two vectors act on the boat from the same starting point. The boat's actual path and speed relative to the ground is the resultant velocity, which is perfectly represented by the diagonal of the parallelogram formed by the boat's velocity and the river's velocity.
5. Why does the diagonal of the parallelogram correctly represent the sum of the two vectors?
The diagonal correctly represents the sum because of the principles of vector translation. In a parallelogram with adjacent sides representing vectors A and B, the side opposite to B is parallel and equal to B. If you slide vector B to this opposite side, you create a triangle where A and the translated B are arranged head-to-tail. According to the Triangle Law, the closing side of this triangle is the resultant A + B. This closing side is exactly the diagonal of the original parallelogram.
6. How is the formula for the magnitude of the resultant vector derived using the Parallelogram Law?
The derivation involves basic geometry and trigonometry. Starting with vectors A and B forming a parallelogram, we extend the line of vector A and drop a perpendicular from the head of vector B to this extended line. This creates a right-angled triangle. By applying the Pythagorean theorem to this larger triangle, and using trigonometric identities (cos θ and sin θ) for the smaller triangle, we can algebraically derive the formula R² = A² + B² + 2AB cos(θ), which gives the magnitude of the resultant.
7. Under what specific conditions is the Parallelogram Law of Vector Addition applicable?
The Parallelogram Law is applicable under two key conditions:
- The vectors must be of the same type (e.g., you can add two force vectors, but not a force vector and a velocity vector).
- The vectors must be co-initial, meaning they must be acting on the same point and at the same time. The law is designed to find the net effect of two forces or quantities originating from a single point.
8. How is the Parallelogram Law used when dealing with forces?
When two forces act simultaneously on an object at a single point, the Parallelogram Law is used to find the net force or resultant force. The two forces are drawn as adjacent sides of a parallelogram, and the diagonal represents the resultant force in both magnitude and direction. This resultant force has the same effect on the object as the two individual forces combined.
