Question

A 5.2 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. How far to the left of the pivot must a 4.0 kg cat stand to keep the seesaw balanced

Answer 1

To balance the seesaw with the given masses and distances, the 4.0 kg cat must stand 0.615 m to the left of the pivot point.

Step-by-step explanation:

To balance the seesaw, we need to consider both the masses and distances from the pivot point. The total torque on the seesaw must be zero for it to be in equilibrium.

The torque equation is given by:

Torque = Force × Distance

Setting up the equation using the given values:

(2.5 kg)(2.0 m) = (5.2 kg)(4.0 m - x)

Simplifying and solving for x:

2.5 kg × 2.0 m = 5.2 kg × 4.0 m - 5.2 kg × x

x = (5.2 kg × 4.0 m - 2.5 kg × 2.0 m) / 5.2 kg

x = 0.615 m

So, the 4.0 kg cat must stand 0.615 m to the left of the pivot point to keep the seesaw balanced.