Mechanical equilibrium- please explain process?

Question

A board 2 m long with a mass of 5 kg is balanced horizontally on Kevin’s shoulder. A 1 kg backpack is hanging from one end of the board. Find the distance between Kevin’s shoulder and the other end of the board. Find the force (magnitude and direction) that Kevin’s shoulder exerts on the board.

Answer 1

In problems of this type where all the forces are parallel, you can:

(a) take moments about any point and equate the total clockwise moment to the total anticlockwise moment,

(b) equate the total upward force to the total downward force.

Let x m. be the distance from Kevin's shoulder to the other end of the board.

The distance from his shoulder to the bag is 5 - x m. and the distance from his shoulder of the centre of mass of the plank is x - 2.5 m.

Taking moments about his shoulder:

5(x - 2.5) = 5 - x

5x - 12.5 = 5 - x

6x = 17.5

x = 2.92 m.

Equating the upward force F of Kevin's shoulder on the plank to the total downward force on the plank:

F = (5 + 1)g

= 6 * 9.81

= 58.9 N.

The force exerted on Kevin's shoulder by the plank is 58.9 N vertically downwards.