Question

A metal bar of 70 cm long and 4 kg in mass supported on two knife edges placed 20 cm from each end. A 6 kg load is suspended at 30 cm from on end. Find the normal reaction at the knife-edge. (assume it to be of uniform cross section and homogeneous).

• A. 55 N at $K_1$ and 33 N at $K_2$
• B. 45 N at $K_1$ and 53 N at $K_2$
• C. 55 N at $K_1$ and 43 N at $K_2$
• D. 43 N at $K_1$ and 55 N at $K_2$

Answer 1

Answer: (C)

55 N at $K_1$ and 43 N at $K_2$

Let the normal reactions from the knife edges $\left(K_1\&K_2\right)$ be $N_1,N_2$.

Now since the bar is in equilibrium,

$N_1+N_2=mg$

$N_1+N_2=10\times 9.8N=98N$

Also the system is in rotational equilibrium too.

Thus, about the center of the rod, the net torque must be zero.

Hence $(6\times 9.8)\left(5\right)+\left(N_2\right)\left(25\right)=\left(N_1\right)\left(25\right)$

$⟹N_1=55N$

$N_2=43N$

Hence normal reaction 55 N at $K_1$ and 43 N at $K_2$ are formed.