Question

A wrestler is going in for a double leg take down against an opponent who weighs 110 kg. the opponent’s center of gravity is 0.8 m above the ground, and the horizontal distance between his opponent’s center of gravity and hind leg is 0.4 m. the wrestler applies a directly horizontal toppling force at a height of 1.2 m. In order to take his opponent down, how large must the toppling force applied to his opponent be?

a. Greater than 719.4 N

b. Greater than 864.4 N

c. Equal to 565.3 N

d. Greater than 359.7 N

Answer 1

The self weight 110kg of the wrestler is acting at the centre of gravity of his body which is at 0.4meters from the hind leg.

The horizontal going force should overcome the moment created by the self weight.

At equilibrium the moments caused by self weight and toppling force must be equal.

Let the toppling force be Ft

Self weight be W

Acceleration due to gravity be g

Then, at equilibrium

Therefore the answer is option D, 359.7N