A metre rule is pivoted at its mid-point with a vertical force of 10N hanging from the distance 30cm from the mid-point. At what distance must a 15N force h...

A metre rule is pivoted at its mid-point with a vertical force of 10N hanging from the distance 30cm from the mid-point. At what distance must a 15N force hang to balance the ruler horizontally?

Answer Details

To balance the ruler horizontally, the 15N force must hang at a distance of 20cm from the mid-point. The idea behind balancing the ruler is to ensure that the torques (or rotational forces) caused by the forces on either side of the pivot point are equal. Torque is calculated by multiplying the force by the distance from the pivot point. In this case, the 10N force creates a torque of 10N * 0.3m = 3Nm at a distance of 30cm from the pivot point. To balance this torque, we need to apply a force of 15N at a distance of x from the pivot point. The torque created by this force must equal 3Nm, so we can set up an equation to solve for x: 15N * x = 3Nm Dividing both sides by 15N, we get: x = 3Nm / 15N = 0.2m = 20cm So, the 15N force must hang at a distance of 20cm from the mid-point to balance the ruler horizontally.