Question

A rod AB of mass 10 kg tied with a string at C such that AC = BC and rod remains in equilibrium then friction force by horizontal surface will be –

• A. 50 N towards right
• B. 50 N towards left
• C. 100 N
• D. None of these

Answer 1

Answer: (A)

50 N towards right

Answer 2

T = f ….(1)

N = mg ….(2)

taking torque about A

T × $\frac{L}{\sqrt{2}}$ = mg × $\frac{L/2}{\sqrt{2}}$

 T = $\frac{mg}{2}$ = 50 N