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?
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
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