Force On an Inclined Plane

  • #1

Homework Statement




A 50 kg block on a frictionless inclined plane experiences a horizontal force. The plane is inclined at 30 degrees.

a- If the block moves up the ramp with a constant velocity, find the magintude of 'F'.
b- Suppose the block accelerates up the ramp at 0.5 m/s^2. Find 'F' now.

Homework Equations



EF = ma...?

The Attempt at a Solution



I drew a diagram with F going left on the x, nx going right on the x, ny going up on the y, and w going down on the y.

Do I need to find a maginitude for the normal force? How can I do that? I believe nx = F and ny = w, but where do I go from there, if that's even correct?
 
Last edited:

Answers and Replies

  • #2
159
0
What is nx and ny and w?


You should resolve the Weight into x and y components, not the normal force.
 
  • #3
w = weight, nx and ny are the components of the normal force, which I drew pointing up perpendicular to the plane. I didn't think I needed two components for the weight, since it is pointing straight down?
 
  • #4
LowlyPion
Homework Helper
3,090
4
w = weight, nx and ny are the components of the normal force, which I drew pointing up perpendicular to the plane. I didn't think I needed two components for the weight, since it is pointing straight down?
The inclined plane separates the force of gravity which is straight down insofar as looking at kinematics on the plane.

The Normal force is perpendicular to the plane. Horizontal force is along the plane. That means the Normal force is M*g*Cosθ The force down the incline is mg*Sinθ .

So what value of F would leave the block in constant motion? (Hint: no net horizontal force along the incline.)

Note when I say horizontal I mean horizontal along the incline. If the force is horizontal along the plane that the incline is on, then that would be different.
 
  • #5
The way I drew it is the only way I've been shown. To make it clear, the force IS parallel to the plane that the incline is on, not to the plane.
 
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  • #6
LowlyPion
Homework Helper
3,090
4
The way I drew it is the only way I've been shown. To make it clear, the force IS parallel to the plane that the incline is on, not to the plane.
Well that makes it simpler then.
 
  • #7
OK, so back to the original question...

Do I need to find a maginitude for the normal force? How can I do that? I believe nx = F and ny = w, but where do I go from there, if that's even correct.
 
  • #8
LowlyPion
Homework Helper
3,090
4
OK, so back to the original question...

Do I need to find a maginitude for the normal force? How can I do that? I believe nx = F and ny = w, but where do I go from there, if that's even correct.
As long as there is no friction you don't need that force. What ever it is the plane pushes back.
 
  • #9
I thought I needed that because the normal force's x component is opposite the F force in the diagram. If not, what do I need?
 
  • #10
73
0
what is the point of finding the magnitude of the normal force.
 
  • #11
If there is no point to finding the normal force, could someone please explain to me how to solve this?
 
  • #12
73
0
ok I think You have to make tow formulas let me give you litle hint
The force in the X- direction will be F - cos(teta) mg = ma
DONT forget the velocity is constant.

and Force in the Y- direction is N - sin(teta)mg = ma
Dont forget the acceleration would be zero.
 
  • #13
Alright. Thanks for helping!
 

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