A load weighing 300 kg is lifted from a 3m long arm by applying a force of 600N to the end of the arm.

A load weighing 300 kg is lifted from a 3m long arm by applying a force of 600N to the end of the arm. At what distance from the center of the arm is the fulcrum? The weight of the load is applied to the other end of the arm, the weight of the arm can be neglected.

Let’s find the force of gravity acting on the load: F = m * g, where m is the mass, g is the acceleration of gravity. Then:

F = 300 * 10 = 3000 N.

The leverage rule says:

F1 / F2 = L2 / L1, where F1 and F2 are forces, L1 and L2 are shoulders.

Obviously, L1 + L2 = L where L is the total length of the arm.

Find the smaller of the shoulders (L2):

L2 = F1 / F2 * L1.

Let us express L1 in terms of L2:

L1 + L2 = L;

L1 = (L – L2).

L2 = F1 / F2 / (1 – F1 / F2) * L.

L2 = 600/3000 * 3 / (1 – 600/3000) = 3/4 m.

Since the distance to the middle is:

3: 2 = 3/2 m.

The difference is:

3/2 – 3/4 = 3/4 m