Question

A uniform rod 1m long of mass 50g is supported horizontally on two knife edge place 10cm from its end. What will be the reaction at this support when 100g mass is suppend 10cm from the midpoint of the rod

Answer 1

Let us consider the forces acting on the uniform rod (see the figure for details):

Applying the equilibrium conditions for the net force and net torque (in resprect to the O-axis), one can write:

$$N_1 + N_2 - mg - Mg = 0\\ mg \cdot 0.4 + Mg \cdot 0.5 - N_2 \cdot 0.8 = 0$$

From the 2nd equation we get:

$$N_2 = \frac{(0.4m + 0.5M)g}{0.8} = \frac{(0.4 \cdot 0.05 + 0.5\cdot 0.1)10}{0.8} = 0.875 \, N \approx 0.86 \, N$$

Substituting this result into the 1st one, we derive:

$$N_1 = (M+m)g - N_2 = 0.15 \cdot 10 - 0.86 = 0.64 \, N$$

Answer: the reaction forces are 0.64 N and 0.86 N correspondingly.