A uniform rod of weight 5N and length 1m is pivoted at the 20cm mark from one end. A weight is hung from the other end so that the rod balances horizontally. What is the value of the weight?
To find the value of the weight, we can set up an equation using the principle of torque balance.
Torque is the tendency of a force to rotate an object about a particular point or axis. It is given by the product of the force and the perpendicular distance from the point or axis to the line of action of the force. In this case, we will consider the pivot point as the axis of rotation.
Let's denote the unknown weight as W.
On one side of the pivot point, there is the weight of the rod, which we know is 5N, acting at a distance of 20cm = 0.2m from the pivot. The torque due to this weight is given by:
torque_rod = 5N × 0.2m
On the other side of the pivot point, we have the weight W acting at a distance of 1m - 0.2m = 0.8m from the pivot. The torque due to this weight is given by:
torque_weight = W × 0.8m
Since the rod is balanced horizontally, the torques on both sides must be equal. Therefore, we can set up the equation:
torque_rod = torque_weight
5N × 0.2m = W × 0.8m
Simplifying the equation, we find:
1N = W
Therefore, the value of the weight W is 1N.