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?

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 be the force exerted by the other player on you.

Horizontal force exerted by other player is where is the angle between the direction of force and horizontal.

Given,

So,the horizontal force is

Vertical force exerted is

Normal reaction force from ground is the sum of vertical force and my weight=

Let the coefficient of static friction be

given,

Maximum frictional force is

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

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west