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Horizontal Force Required to move a box up a ramp

  • Thread starter Bahrbarian
  • Start date
  • #1

Homework Statement


A block with a mass of m is initially moving up the incline and is increasing speed with an acceleration of a. The applied force F is horizontal. The coefficients of friction between between the block and incline are μs=S and μk=K. The angle of the incline is θ.
What is the magnitude of F?


Homework Equations


I know Im using f=ma. I just am confused as to how all the forces are found, and ofr their directions, and how to find the force of friction.


The Attempt at a Solution


I left all of the terms in just variables so I would learn how to solve this, not just the answer. I have been doing this on some online homework, so I can check my answer, and I don't understand what I am doing wrong. Magnitude of the normal force is equal to F+mg, I think, and then I split that into horizontal and vertical components. And the force of friction should be SxN, which would be Sx(F+mg), I think? I have to eventually calculate what F equals, and what the force of friction equals. I also think that the coefficient for static friction is extraneous information, but Im not positive. Any help would super awesome.
 

Answers and Replies

  • #2
Bandersnatch
Science Advisor
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Fg=mg is NOT normal to the incline. Neither is F.
The latter, as said in the problem description, is horizontal. The former... well, think about the direction gravity should be pulling the block. If it were normal, it'd mean that we could walk on walls.
Now take these two forces, and find their normal and parallel to the incline components.
The parallel ones will be responsible for the body moving, the normal ones for the magnitude of the force of friction that opposes the movement(the total normal force times the friction coefficient).
If you have problems with splitting the forces into components correctly, remember that you're building a rectangle with the original force as a diagonal.
You can relate the component forces to mg and F via trigonometry.
 
Last edited:
  • #3
I know that F and mg are not not normal to the incline. F is horizontal, mg is vertical. But youre saying that the normal force isnt equal to F+mg? If not, itd be like Fxsinθ+mgcosθ, yes? The magnitude of the normal force, perpendicular to incline, youre saying isnt F+mg?
 
  • #4
Bandersnatch
Science Advisor
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If not, itd be like Fxsinθ+mgcosθ, yes?
You've got it.
Do the same for parallel forces, add friction to the mix, and see if you can find the answer.
 

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