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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?

User HUSTEN
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1 Answer

2 votes

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.

User Cintu
by
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