Force needed to push box down incline plane

AI Thread Summary
To determine the force needed to push a 120N block down a 20-degree incline at constant speed, the friction force and the component of weight along the incline must be considered. The applied force to push the block up is 86N, which balances the weight component and friction. The coefficient of friction was calculated to be approximately 0.399. For the block to move down at constant speed, a force of about 3.9N is required. This analysis confirms the relationship between applied force, friction, and weight on an incline.
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Homework Statement


a) if a force of 86N parallel to the surface of a 20 degree incline plane will push a 120N block up the plane at constant speed, what force parallel to the plane will push it down at constant speed?


Homework Equations


\SigmaF(X)=F(A)-F(fr)-mgcos\vartheta=0
\SigmaF(Y)=F(n)-mgsin\vartheta=o
where F(A)= force applied, F(fr)= friction force, F(n)= normal force, mg=weight, F(X)= sum of forces in x direction, and F(Y)- sum of forces in y direction


The Attempt at a Solution


normal force=41.0424
friction force=26.73
my answer is -86.0331
 
Last edited:
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my coefficient of friction for this problem=.652
 
could someone please check my answers?
 
Physics_girl1 said:

Homework Statement


a) if a force of 86N parallel to the surface of a 20 degree incline plane will push a 120N block up the plane at constant speed, what force parallel to the plane will push it down at constant speed?


Homework Equations


\SigmaF(X)=F(A)-F(fr)-mgcos\vartheta=0
\SigmaF(Y)=F(n)-mgsin\vartheta=o
where F(A)= force applied, F(fr)= friction force, F(n)= normal force, mg=weight, F(X)= sum of forces in x direction, and F(Y)- sum of forces in y direction


The Attempt at a Solution


normal force=41.0424
friction force=26.73
my answer is -86.0331

Physics_girl1 said:
my coefficient of friction for this problem=.652

Physics_girl1 said:
could someone please check my answers?

Hi
Writing out the equation for the force to push up the block with constant speed, i.e. net force is zero.
120 sin 20(component of weight along incline) +120 cos 20 * u(Force of friction, u is the coefficient)=86
solve for u
u=0.399
Now for the second part:
120 sin 20 +F=120 cos 20 *u
F=120 sin 20-120 cos 20 *u
F=3.9 N
dos this help?
 
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