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 'Î¸' 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.
Observations and Readings
i. To find the weight of the body by measuring the diagonal
Scale chosen 1 cm = 50 g.wt
Trial 
Forces 
OA 
OB 
OD 
W_{1} = OD x scale factor 
Mean W_{1} 

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 (W_{1}) = 180 g.wt = 180 x 10^{3} kg.wt.
ii. To find the weight of the body by measuring the angle
Trial 
Forces 
Î˜ = ∠XOY 
W_{2} 
Mean W_{2} 

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, W_{2} = 174.30 g.wt = 174.30 x 10^{3} kg.wt
Mean weight, W = (W_{1} + W_{2})/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 = W_{a}/(W_{a} – W_{w})
ii. Relative density of liquid = Loss of weight of the solid in liquid / Loss of weight of the solid in water = (W_{a} – W_{l})/(W_{a} — W_{w})
where W_{a} = Weight of the body in air
W_{w} = Weight of the body in water
W_{l} = Weight of the body in liquid
Procedure
The weight of the given body in air (W_{a}) 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 (W_{w}). Then the given body is immersed in liquid and the weight (W_{l}) is determined as mentioned above. Then relative density of the given solid and liquid can be calculated.
Observations and Readings
Body in 
Trial 
P 
Q 
OA 
OB 
OD 
Weight of body 
Mean 
Air 
1 
100 
100 
2 
2 
1.7 
92.5 
W_{1} = 92.5 
2 
150 
150 
3 
3 
1.6 
92.5 

Water 
1 
100 
100 
2 
2 
1.4 
70 
W_{2} = 70 
2 







Liquid 
1 
100 
100 
2 
2 
1.3 
65 
W_{3} = 65 
2 






Weight of the body in air, W_{a} = 92.5 gwt
Weight of the body in water, W_{w} = 70 gwt
Weight of the body in liquid, W_{l} = 65 gwt
Relative density of the solid = W_{a}/(W_{a}  W_{w}) = 4.16
Relative density of the liquid = W_{a}W_{l}
/ W_{a}W_{w} = 1.2
Results
Relative density of the solid = 4.16
Relative density of the liquid = 1.2
Density of the Solid = 4.16 x 10^{3} kgm^{3}
Density of the liquid = 1.2 x 10^{3} 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.
For More Viva Questions, see Viva Questions on Parallelogram law of forces
About Sreejith Hrishikesan
Sreejith Hrishikesan Nair is a MTech graduate in Communication Systems. He completed Btech Degree in Electronics and Communication.He is a person who wants to implement new ideas in the field of Technology.
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