A pole AB of length 10m and weight 600N has its centre of gravity 4m from the end A,and lies on horizontal ground. Calculate the force required to beging tto lift the end. Why would this force applied at the end A not be sufficient to lift the end A
a pole ab 10cm long and weighing 600n has its centre of gravity 4cm from side a linning on the horizontal ground show that the force required to lift side b is not enough to lift side a
To calculate the force required to begin lifting the end of the pole, let's consider the equilibrium condition first. When the pole is in equilibrium, the sum of all the forces acting on it must be zero.
Since the pole lies on horizontal ground, there are two forces acting on it:
1. The weight of the pole acting vertically downward, which is 600N.
2. The normal force of the ground pushing upward to balance the weight, which is also 600N.
Now, to lift the end A, we need to overcome the downward force of the weight acting at the center of gravity (4m from end A). To do this, we need to apply an upward force at the end A that is greater than the weight force.
If we apply a force exactly equal to the weight force at the end A, the net force would be zero, and the pole would remain in equilibrium. It would not start lifting from end A because the opposing forces would cancel each other out.
To lift the end A, we need to apply an additional force greater than the weight force. This additional force creates an unbalanced force, resulting in a net force pointing upward, causing the pole to lift from end A.
So, applying a force equal to the weight force at the end A would not be sufficient to lift the end A because it would only create an equilibrium situation, where the net force is zero. To start lifting the end A, we need to apply an upward force greater than the weight force at the end A.