Answer to Question #146600 in Mechanics | Relativity for emily

Question

A man carries a load of mass 2.6kg from one end of a uniform pole 100cm long which has a mass 0.4kg. The pole rest on his shoulder at a point 60.0cm from the load and he holds it by the other end. What vertical force must be applied by his hand and what is the force on his shoulder?

Answer 1

The system consists of a pole with 2.6 kg load at one end. Lets say it is creating clockwise torque.

The uniform pole has its own weight considered at centre of the pole, which also generates clockwise torque.The hand will have to provide counterclockwise torque to balance the situtaion. 

The shoulder acts as the pivot point

net τ = 0

counterclockwise torque - clockwise torque = 0

F(hand)(0.400 m) - [(2.6 kg)(9.8 $\tfrac{m}{s^2}$ )(0.600 m) +(0.4 kg)(9.8 $\tfrac{m}{s^2}$ )(0.100 m)] = 0

$F(hand)=$ $\dfrac{[(2.6 kg)(9.8 \tfrac{m}{s^2} )(0.600 m) +(0.4 kg)(9.8 \tfrac{m}{s^2} )(0.100 m)]}{(0.400 m)}$

$F(hand) = 39.2 N$ ← vertical force to be applied by the hand

Force on the shoulder = net vertical load = F(hand) + weight of pole + weight of load

F(shoulder) = (39.2 N) + (0.4 kg)(9.8$\tfrac{m}{s^2}$ ) + (2.6 kg)(9.8 $\tfrac{m}{s^2}$ )

F(shoulder) = 68.6 N downwards