Equilibrium Rotational Motion

Question

A 10.5 kg board 6.00 m long is

supported by two sawhorses, one

at each end.

A 4.45 kg saw sits on

the board 1.80 m from the left end.

Find the upward force that the

LEFT sawhorse exerts on the

board

(Unit = N)

Answer 1

Let,

M = 10.5 kg be the mass of the board,

m = 4.45 kg be the mass of the saw,

D = 6 m be the length of the board,

d = 1.8 m be the distance of the saw from the left end,

g = -9.81 m/s² be gravitational acceleration,

F (to be found) be the force of the left sawhorse.

I will treat upward direction as positive and downward direction as negative;

that is why the gravitational acceleration is negative.

If the left sawhorse collapse, the board would rotate around the right end.

This rotation would be caused by the torque of the saw and the board itself.

In contrast, in the presence the force F of the left sawhorse, there is no rotation.

Therefore the left sawhorse causes the the sum of all the torques to be 0.

If we assume the center of the board's gravity to be in the middle,

then the board acts with force Mg, distance D/2 from the right end.

The saw acts with force mg, distance D-d from the right end.

The left sawhorse acts with force F, distance L from the right end.

Thus the condition for the sum of all the torques to be 0 becomes

MgD/2 + mg(D-d) + FD = 0

From that

F = -g(M/2 + m(1 - d/D))

Substituting actual numbers

F = 9.81×(10.5/2 + 4.45×(1 - 1.8/6)) = 82.1 N