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HELP: Friction on upward motion in an inclined plane

  • #1
Guys, I'm just bothered of my professor's answer in the problem we solved. Here are the given

Given:
Weight: 5000 N
Coefficient: 0.4
Angle of the plane: 20º

Solve for the required FORCE to move the box (5,000N) upward to the plane.

This is the formula I used:

Fn=Ff

Fn(sin theta) = (coefficient)(5000 N - normal force)(cos theta)

Fn = (coefficient)(5000 N - normal force)(cos theta)/sin theta

Fn = 5, 494.95 N

Am I right or wrong?

Thanks for the help.
 

Answers and Replies

  • #2
24
0
I may be wrong but i think of it this way;

Force down the plane by gravity is 5000*sin(20)=1710.1N
Force of friction is 5000*cos(20)*0.4=1879.4N

A force pushing it up the plane needs to overcome both of these forces, so the force needs to be about 1710.1+1879.4=3589.5N

I think thats right. The force you calculated is actually larger than the weight of the box, which i think is intuitively too large
 
  • #3
15
0
The angle of plane is given 20degree
the force is resolved into two component that is
1.Horizontal component
formula
fx=cos(angle)f1
2.Vertical component
fy=sin(angle)f1
Determine the net force here is the formula
total force=sqrt(fx^2+fy^2)
this is the exact solution
 
  • #4
I may be wrong but i think of it this way;

Force down the plane by gravity is 5000*sin(20)=1710.1N
Force of friction is 5000*cos(20)*0.4=1879.4N

A force pushing it up the plane needs to overcome both of these forces, so the force needs to be about 1710.1+1879.4=3589.5N

I think thats right. The force you calculated is actually larger than the weight of the box, which i think is intuitively too large
This is the correct answer, as per my professor. Thanks for the light.
 

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