Crate on a frictionless ramp, constant speed.

[Koscher]

Homework Statement

A 130 kg crate is pushed at constant speed up the frictionless 23° ramp shown in the figure. What horizontal force F is required?

Homework Equations

F=ma
F(parallel)=mgsin(theta)
F(perpendicular)=mgcos(theta)

The Attempt at a Solution

I know that because it is at a constant speed the net force equals zero, therefore the force caused by gravity equals the applied force. So first I solved the parallel force caused by gravity, which equals 498.2994 N. From there to create a net force of zero I flipped the sign, -498.2994 N. Solved for the horizontal component of that force: -498.2994cos23 =-458.687 N.

But obviously that was not right. I think I might have to include the normal force by gravity in that equation too though. Any thoughts?

 

[Doc Al]

I know that because it is at a constant speed the net force equals zero, therefore the force caused by gravity equals the applied force. So first I solved the parallel force caused by gravity, which equals 498.2994 N. From there to create a net force of zero I flipped the sign, -498.2994 N. Solved for the horizontal component of that force: -498.2994cos23 =-458.687 N.

Since the applied force is horizontal, you have found the component of F parallel to the incline. Now you need to find the full force F. Express that mathematically so you can solve for F.

 

[Koscher]

I am afraid that I am a little lost at the moment. I need to express the full force of the equation?

 

[Doc Al]

If the applied force F is horizontal, what would be its component parallel to the incline?

 

[Koscher]

Force applied(parallel) = Force applied(sin(theta))

So: 498.2994/sin(23) = 1275.3 N [force applied]

 

[Doc Al]

Force applied(parallel) = Force applied(sin(theta))

Not exactly. Realize that the applied force is horizontal, not vertical (like gravity).

 

[Koscher]

Okay so the applied force in the direction of parallel, would that be an equal but opposite force to the force parallel caused by the force of gravity?

 

[Doc Al]

Okay so the applied force in the direction of parallel, would that be an equal but opposite force to the force parallel caused by the force of gravity?

Yes, if I understand what you mean. The net force in any direction must be zero. Thus the forces parallel to the ramp must be zero. Thus the parallel component of gravity and the parallel component of the applied force must be equal and opposite.

 

[Koscher]

Okay.

If I take the horizontal component of the Normal force and the horizontal component of the applied force and add them together, will that give me a total horizontal force.

(498.299cos23)=458.6866 N
(1380sin23)=539.208 N

So, 997.8946 N?

 

[Doc Al]

If I take the horizontal component of the Normal force and the horizontal component of the applied force and add them together, will that give me a total horizontal force.

Since the crate is in equilibrium, the net horizontal force will be zero. Some problems with this approach:
- The applied force is totally horizontal, so the horizontal component of the applied force is the applied force
- The normal force does not simply equal mgcosθ

You'll have a much easier time if you stick to the force components parallel to the ramp.

 

[Koscher]

Okay.

So.

Fcos(theta)-mgsin(theta)=ma
Fcos(23)-(130kg)(9.81 m/s^2)sin(23)=0 (it equals zero because of the zero net force)

Solve for F.

F= 541.332 N

Sorry for being a pain, I have been fine with similar questions, just this one was confusing. You were a huge help, thank you. That would be the answer if I am not mistaken.

 

[Doc Al]

Good!