Question

let R and S be the resultant of two forces P and Q acting at a point when the angle between two forces is 60 and 120 respectively. show that R^2 + S^2= 2( P^2 + Q^2)​

Answer 1

The proof is given below:

We know that resultant R of two forces P and Q acting at a point at an angle $\theta$ is given by

$\boxed{R=\sqrt{P^2+Q^2+2PQ\cos\theta}}$

According to the question

If $\theta=60^\circ$

$R=\sqrt{P^2+Q^2+2PQ\cos 60^\circ}$

$\implies R^2=P^2+Q^2+2PQ\times \frac{1}{2}$

$\implies R^2=P^2+Q^2+PQ$     .......... (1)

Similarly, when the forces are acting at an angle $\theta=120^\circ$

$S^2=P^2+Q^2+2PQ\cos 120^\circ$

$\implies S^2=P^2+Q^2+2PQ\cos (180^\circ-60^\circ)$

$\implies S^2=P^2+Q^2-2PQ\cos 60^\circ$

$\implies S^2=P^2+Q^2-2PQ\times \frac{1}{2}$

$\implies S^2=P^2+Q^2-PQ$  ........... (2)

Adding equation (1) and (2)

$R^2+S^2=(P^2+Q^2+PQ)+(P^2+Q^2-PQ)$

$\implies R^2+S^2=2(P^2+Q^2)$

Hope this answer is helpful.

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Q: let R and S be the resultant of two forces P and Q acting at a point when the angle between two forces is 60 and 120 respectively. show that R^2 + S^2= 2( P^2 + Q^2)​

Click Here: https://brainly.in/question/30014

Answer 2

Given : R and S be the resultant of two forces P and Q acting at a point when the angle between two forces is 60 and 120 respectively

To Find : show that R^2 + S^2= 2( P^2 + Q^2)​

Solution:

R is resultant force of P & Q  when angle between them is 60°

Hence R² = P² + Q² + 2PQCos60°

Cos60° = 1/2

=> R² = P² + Q² + 2PQ(1/2)

=> R² = P² + Q² +  PQ

S is resultant force of P & Q  when angle between them is 120°

Hence S² = P² + Q² + 2PQCos120°

Cos120° = -1/2

=> S² = P² + Q² + 2PQ(-1/2)

=> S² = P² + Q² -  PQ

Adding both

R²  + S² =  P² + Q² +  PQ + P² + Q² -  PQ

=> R²  + S² =  2( P² + Q² )

QED

Hence proved

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