Question

You are 80 kg and you want to push over a player that is 100 kg with a force of 1200 N at 20 degrees above the horizontal. You both have a static friction coefficient of 0.95. If he is pushing back with 1200 N at 20 degrees below horizontal on you, will you accelerate backwards?

Answer 1

Answer:I will not accelerate backwards.

Step-by-step explanation:

Whenever we want to discuss the motion of a body,we need to consider the forces acting on that body and not the force it is exerting on other bodies.

Let
`F` be the force exerted by the other player on you.

Horizontal force exerted by other player is
`FCos(\alpha )` where
`\alpha` is the angle between the direction of force and horizontal.

Given,

`F=1200N\\\alpha =20^(0)`

So,the horizontal force is
`F_(h)=1200* Cos20=1127.63N`

Vertical force exerted is
`F_(v)=1200* Sin(20)=410.42N`

Normal reaction force from ground is the sum of vertical force and my weight=
`N=80(9.8)+410.42=1194.42N`

Let the coefficient of static friction be
`c_(s)`

given,
`c_(s)=0.95`

Maximum frictional force is
`f=N* c_(s)=1194.42* 0.95=1134.7N`

Since the maximum frictional force is greater than the horizontal force,I will not accelerate backwards.