Question

A crowbar 28 in. Long is pivoted 6 in. From the end. What force must be applied at the end in order to lift a 400-lb object at the short end?

Answer 1

What you are trying to do is balance the “moments" about the fulcrum (pivot).

We will calculate moment at the pivot (M1) due to weight (W):

• L1 = length of the bar = 6in

`M1=W* L1=400\cdot6=2400in\cdot lb_f`

The moment (M2) at the pivot due to your applied force (Fa) on the other end of the bar must equal M1.

• LT = total lenght = 28in

,

• L2 = LT - L1 = 28 - 6 = 22in

,

• M2 = M1 = 2400in lbf

`\begin{gathered} M2=Fa* L2 \\ Fa=(M2)/(L2)=(2400inlb_f)/(22in)=109.09lb_f \end{gathered}`

Answer: 109.09lbf

A force of 109.091 pounds would have to be applied to move the load.