A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 50 cm from the end of the plank with force F2. If the plank has a mass of 20 kg and its center of gravity is at the middle of the plank, what are the forces F1 and F2?

F1 = 196 N

F2 = 392 N

Explanation:

Given:

length of the plank = 2 m

mass of the plank = 20 kg

Weight of the plank = 20 x 9.8 = 196 N

Torque due to the weight of the plank with respect to the pivoted end (i. e the end held by the hand) Counter clockwise torque = 196 x cog of wood

= 196 x 1 = 196 Nm

Clockwise torque = F2 x 0.5

for the balanced case

F2 x 0.5 = 196

F2 = 196 / 0.5

F2 = 392 N

now,

the net force

Net downward force = Net upward force

F1 + weight of plank = F2

F1 + 196 = 392N

F1 = 392 - 196

F1 = 196 N