Question

A 5.2 m long, 4 kg board is laid down so that the end of the board extends 1.7 m beyond the end of a pier. Somebody has placed a 0.25 kg can of sardines at the end of the board to lure out a cat. Imagine! How far from the edge of the pier can the 5 kg cat go before the board falls into the water?

Answer 1

O is the end of the pier

AB is the board of length 5.2 m and of 4 kg

AC = CB = 2.6 m

The cat will start moving from the point A

The can of sardines is placed at point B

OB = 1.7 m

OC = 0.9 m

The board will falls into the water when the torque about point O will be larger on the side of the can

The net torque about the point O due to the board and the can is :

OC*Weight of the board(anti clockwise) + OB*Weight of the can (clockwise)

4*g*0.9 - 0.25*g*1.7 (net torque will be in the anti clockwise direction)

Net torque. = 3.175*g (anti clockwise)

After the car starts moving, the board will falls of if a torque of 3.175*g will be generated by the cat in the clockwise direction

Hence the cat should cross the point O in order to generate a clockwise torque

Let the car move a distance x crossing the point O, hence torque of the cat will be 5*g*x

This torque of cat will be equal to 3.175*g

Hence, 3.175*g = 5*g*x

We get, x = 0.635 m

Hence the total distance the cat moves = 2.6 + 0.9 + 0.635 = 4.135 m