Answer to Question #141943 in Physics for leen

Question

On a see-saw mary, weight 600N balances john, weight 200N when she sits 1.5m away from the pivot.how far from the pivot is john?

Answer 1

From the torques equality, we can write:

$$600N\times 1.5m = 200N\times x$$

where $x$ is the distance from the pivot to John. Expressing $x$, find:

$$x = \dfrac{600N\times 1.5m}{200N} =4.5m$$

Answer. 4.5 m.