Concurrent Forces Experiment

Concurrent Forces Experiment for Class 11 with Readings

EXPERIMENT 1 : CONCURRENT FORCES - I

Aim

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

Apparatus

A vertical wooden board, Weight hangers, Slotted weights, Pins, Sheet of paper etc.

Principle

Parallelogram law of vectors states that if two adjacent sides of a parallelogram represent two given vectors in magnitude and direction, then the diagonal of the parallelogram starting from the point of intersection of the two vectors represent their vector sum.

Let P and Q be the magnitude of two forces (vectors) acting at a point and 'θ' be the angle between them. Then magnitude of the resultant,

Procedure

i. To find the weight of the body by measuring the diagonal

Three light inelastic strings are tied together to form a fine knot at O. The two strings which pass over the pulleys carry the weights P and Q. The given body of unknown weight W is tied to the third string which hangs freely. The weights P and Q are so adjusted that the point O is in equilibrium. A sheet of paper is fixed on the drawing board behind the string. The directions OX, OY and OZ of the strings are marked on the paper. The paper is taken out. The lines OX, OY, and OZ are drawn to represent P, Q and W in direction. The experiment is repeated atleast thrice by changing the values P and Q.

Taking suitable scale, distance OA and OB are marked on the lines OX and OY to represent the forces P and Q in magnitude and direction. With OA and OB as adjacent sides, the parallelogram OADB is completed. The diagonal OD is drawn. The length of OD is measured. It is multiplied with the scale. This gives the weight  of the body.

(ii) To find the weight of the body by measuring the angle

The angle XOY between the forces P and Q is measured. The weight of the body is also calculated using the formula,

The experiment is repeated by varying the values of the weights P and Q. The mean value of the weight of the body is determined.

i. To find the weight of the body by measuring the diagonal

Scale chosen 1 cm = 50 g.wt

 Trial Forces OA OB OD W1 = OD x scale factor Mean W1 P Q 1 100 100 3 3 3.9 50x3.9 = 195 180 2 150 150 4 4 3.5 175 3 200 200 5 5 3.5 175 4 250 250 6 6 3.5 175

Weight of the given body by measuring diagonal (W1) = 180 g.wt = 180 x 10-3 kg.wt.

ii. To find the weight of the body by measuring the angle

 Trial Forces Θ = ∠XOY W2 Mean W2 P Q 1 100 100 90° 141.43 174.30 2 150 150 124° 168.62 3 200 200 136° 192.20 4 250 250 149° 194.96

Weight of the given body, W2 = 174.30 g.wt = 174.30 x 10-3 kg.wt

Mean weight, W = (W1 + W2)/2 = 0.17715 kg.wt

Result

Weight of the given body = 0.17721 kg.wt

EXPERIMENT 2 : CONCURRENT FORCES - II

Aim

To determine the relative density of a solid and liquid by applying parallelogram law of forces.

Apparatus

A vertical wooden board with two pulleys, Weight hangers, Slotted weights, Pins, Sheet of paper, the given body and liquid etc.

The parallelogram law apparatus consists of a vertical drawing board with two smooth pulleys fixed at its top.

Principle

i. Relative density of solid = Weight of the solid in air / Loss of weight in water  = Wa/(Wa – Ww)

ii. Relative density of liquid = Loss of weight of the solid in liquid / Loss of weight of the solid in water = (Wa – Wl)/(Wa — Ww)

where Wa = Weight of the body in air

Ww = Weight of the body in water

Wl = Weight of the body in liquid

Procedure

The weight of the given body in air (Wa) is determined as mentioned in the above experiment. Now the given body is immersed in water taken in beaker without touching the bottom of the beaker. The weight of the body is determined (Ww). Then the given body is immersed in liquid and the weight (Wl) is determined as mentioned above. Then relative density of the given solid and liquid can be calculated.

 Body in Trial P Q OA OB OD Weight of body Mean Air 1 100 100 2 2 1.7 92.5 W1 = 92.5 2 150 150 3 3 1.6 92.5 Water 1 100 100 2 2 1.4 70 W2 = 70 2 Liquid 1 100 100 2 2 1.3 65 W3 = 65 2

Weight of the body in air, Wa = 92.5 gwt

Weight of the body in water, Ww = 70 gwt

Weight of the body in liquid, Wl = 65 gwt

Relative density of the solid = Wa/(Wa - Ww) = 4.16

Relative density of the liquid = Wa-Wl / Wa-Ww = 1.2

Results

Relative density of the solid =  4.16

Relative density of the liquid =  1.2

Density of the Solid = 4.16 x 103 kgm-3

Density of the liquid = 1.2 x 103 kgm-3

MODEL VIVA VOCE QUESTIONS AND ANSWERS

1. Explain the term, force?

Force is that which changes or tries to change the state of rest or of uniform motion of a body.

2. What is meant by composition of vectors?

The process of adding two or more vector quantities to find their resultant.

3. What is the relation between newton and kg wt?

1 kg wt = 9.8 newton

4. Define parallelogram law of forces.

If two vectors (forces) are represented in magnitude and direction along the adjacent sides of a parallelogram, then their resultant can be represented in magnitude and direction by the diagonal drawn from the point of intersection.

5. What is meant by equilibrium?

When a number of forces acting on a body and the body does not change its state of rest or of uniform motion, it is said to be in equilibrium.

6. If there is a lot of friction at the pulleys, in what way the result is affected?

Weight will be lower than the actual value.

7. What is meant by a vector quantity?

Quantities having both magnitude and direction.

8. Give some quantities which are vectors.

Displacement, Velocity, Acceleration, Force

9. What is meant by a scalar quantity?

Quantities having only magnitude and no direction are called scalars.

10. Give some quantities which are scalar.

Mass, Time, Speed, Work

11. Define concurrent forces

Forces whose lines of action meeting at a single point are called concurrent forces.

12. Define coplanar forces.

Forces acting in a plane are called coplanar forces.