# Homework Help: An inclined plane, fixed to the inside of an elevator, makes a 32 degrees angle

1. Nov 18, 2014

### hitemup

1. The problem statement, all variables and given/known data

An inclined plane, fixed to the inside of an elevator, makes a 32 degrees angle with the floor. A mass m slides on the plane without friction. What is its acceleration relative to the plane if the elevator accelerates upward at 0.50g?

2. Relevant equations

F = ma

3. The attempt at a solution

The free body diagram I have drawn involves FN and mg.
I made the relationship between these two like this:

FNcos32 - mg = 0.5mg (upward acceleration)
FNsin32 = m*ax

I've found that ax is 1.5g(tan32).
There is also the acceleration of elevator. But, because we are looking with respect to plane, we don't take it into consideration, do we?

I know that I can simply get the acceleration parallel to the plane just by adding the elevator's acceleration to gravity and multiplying it with sin32. Our teacher, however, doesn't want us to do these problems like that. So I am asking for help about my solution.

Last edited: Nov 18, 2014
2. Nov 18, 2014

### BiGyElLoWhAt

First off, if 32 degrees is the angle between the plane and the floor, do you want F_n cos 32 for the vertical component? (as well as F_n sin 32 for the horizontal?) Second of all, if the mass is on a tilted plane, the weight force on the mass is not equal to mg.

Also, the normal force can essentially be broken down into 2 contributions, did you take both of these into account when you made your calculations?

3. Nov 18, 2014

### hitemup

Yes, FNcos32 is my vertical component, and since the mass is accelerating with elevator, I have written F= ma simply for the mass, that's why I chose to break FN rather than mg.
I'm sorry but can you be more clear on your second sentence?

F_Ncos32-mg=ma_elevator, F_Nsin32=ma_x.
This is all I've done. I believe this must satisfy newton's second law.
From here you get a_x = 1.5gtan32

4. Nov 18, 2014

### BiGyElLoWhAt

Ok so your y axis lies along the plane? And your x axis is into (normal to) the plane
The Normal force is generally mgcos theta. Thats for a plane that is static in the frame of reference. You could do that but it would be m"g"cos theta, which you said is not what your teacher wants you to do. I think you could probably get away with kinematics here, since there is no friction. Find the equation of motion for the plane, find the equation of motion for the block, and take the difference (you'll want to use something static like the elevator shaft as your reference). It's a pretty similar thing to summing the accelerations, though.

Also, you still have to break down mg. It seems as though you're being inconsistent with your coordinate systems. You're using one coordinate system for gravity, and another for the normal. You can't do that.

5. Nov 18, 2014

### hitemup

Let me make it clear that my teacher is not obsessed with not breaking down mg into components. He is just saying that "since it is no real force, you cannot draw ma_elevator next to mg in your free body diagram"

Here I drew my free body diagram.
http://www.sketchtoy.com/63667913

Last edited: Nov 18, 2014
6. Nov 18, 2014

### BiGyElLoWhAt

you can't draw ma_elevator, but you can most definitely draw F_n next to your mg in your FBD. also, I can't view the picture. Can you just post it here in pf?

7. Nov 18, 2014

### hitemup

I said that a_x = 1.5gtan32
If you break a_x into components in my diagram, you can see that it's "cos" component is equal to (1.5gsin32). This is also what you find exactly with your free body diagram: mgsin32+ma_elevatorsin32 = ma)

Last edited: Nov 18, 2014
8. Nov 18, 2014

### BiGyElLoWhAt

Oh man, I'm really sorry about that. The whole time I was saying F_n and thinking F_w.
I still think the kinematics is the way to go. If you sum the forces on the block, you can get the equation of motion for the block. You already know the equation of motion for the ramp.

9. Nov 18, 2014

### hitemup

No problem with that. I'm gonna give it a try with kinematics.
About finding the relative acceleration, we should neglect a_elevator since the plane and mass both have it, right?

10. Nov 18, 2014

### BiGyElLoWhAt

I probably wouldn't, but it should end up dropping out.

11. Nov 18, 2014

### hitemup

This whole inertial and non inertial frame of reference thing messes this up.
In my calculations, I didn't have fictious forces involved, because I decided to choose an inertial frame of reference. But if I were to choose a non-inertial frame of reference, like the plane in elevator(the question was actually asking it as well), I could draw fictious forces, and that would be easier than what I've done, no doubt.
But since my teacher wants no fictious forces, I am obliged to to it in my way. So can I ask if my calculations, especially for a_x, are correct?

12. Nov 19, 2014

### BiGyElLoWhAt

I got a difference of approximately .16 m/s from your answer. I did not round, I used 9.8 for g.

13. Nov 19, 2014

### BiGyElLoWhAt

I applied superposition, eliminated gravity, and used that situation to model how the .5g acceleration affects the mass. Then I put a stationary ramp on the ground in earths gravitational field, and modeled how that affects the block. I then summed them together. Post your work and I'll look a little closer.