A pole AB of length 10m weighs 800N and has its centre of gravity 4m from the end A, and lies on horizontal ground. The least vertical force required to lift its end B is?

f * 10 m = 800 N * 4 m

Answers

To determine the least vertical force required to lift the end B of the pole, we need to consider the torque acting on the pole.

Torque, denoted by τ, is the product of force and the perpendicular distance from the point of rotation (fulcrum) to the line of action of the force. In this case, the fulcrum is the end A of the pole.

We know that the weight of the pole acts vertically downward and can be considered as a force of 800N at the center of gravity, which is 4m from the end A. So, the weight force creates a torque about the end A.

We can calculate the torque by using the formula:

τ = force × distance

The torque created by the weight force of the pole can be calculated as:

τ = 800 N × 4 m
τ = 3200 Nm

For the end B to be lifted, a vertical force must be applied at B that creates a torque opposite to the torque created by the weight force. This is necessary to achieve rotational equilibrium.

So, the magnitude of the minimum vertical force required to lift the end B should be equal to the torque created by the weight force of the pole. Therefore, the least vertical force required to lift the end B of the pole is 3200 N.