Question

A uniform meter rule of weight 1.0N, is pivoted at the 0.40m mark, A2.0N weight is hung from the 0.15m where must a 2.0N weight be placed to balance the meter rule

Answer 1

To balance the meter rule, the torque on each side of the pivot point must be equal. Torque is calculated by multiplying the force applied by the distance from the pivot point.

On one side of the pivot point, we have a 1.0N weight at a distance of 0.40m. So, the torque on this side is 1.0N * 0.40m = 0.40Nm.

On the other side, we have a 2.0N weight at a distance of 0.15m. Let's denote the distance from the pivot point where the 2.0N weight must be placed as "x". The torque on this side would be 2.0N * (0.15m - x).

Since the meter rule is balanced, the torque on both sides must be equal. Therefore, we can set up the equation:

0.40Nm = 2.0N * (0.15m - x)

Now solve for "x":

0.40Nm = 0.30N - 2.0Nx

0.40Nm + 2.0Nx = 0.30N

2.0Nx = 0.30N - 0.40Nm

2.0Nx = 0.30N - 0.40N(0.40m)

2.0Nx = 0.30N - 0.16N

2.0Nx = 0.14N

x = 0.14N / (2.0N)

x ≈ 0.07m

Therefore, the 2.0N weight must be placed at around the 0.07m mark to balance the meter rule.