Question

Forces of 3N, 5N and 7N acting on a particle are in equilibrium, find the angle between the forces 3N and 7N.

Answer 1

Explanation:

Asgiven forces are equilibrium i.e. force of 3N and 7N would be balancing by 5N

so,

We Know That

${r}^{2} = {p}^{2} + {q}^{2} + 2pq \cos$

So,

${5 }^{2} = {3}^{2} + {7}^{2} + 2(3)(7) \cos$

25 = 9 + 49 + 42 cos

cos = -33/42

$theta = { \cos }^{ - 1} (- 33 \div 42)$