Question

Two men are carrying on their shoulders a 500-N load suspended from a piece of bamboo 3m long with each man positioned at the ends. If the load is suspended 1m from one end find the load carried by each.

Answer 1

Given,

Total load = 500N

Length of the bamboo = 3m long

Load is suspended 1 m from one end.

So, $N_1+N_2= 500N ...(i)$

Now applying torque about the one end.

$N_1\times 0 - 500\times 1+N_2\times 3=0$

Hence, $N_2= \frac{500}{3}N$

$N_1=500-\frac{500}{3}N$

$\Rightarrow N_1 =\frac{1000}{3}N$