# Homework Help: Mass on frictionless Plane.

1. Jan 7, 2005

### scuddman

A crate of mass M is placed on a frictionless inclined plane of Angle theta.
Determine the acceleration after the crate is released.

|\
|..\o crate of mass m
|....\
|___b\ angle b which is theta.

I know how to do this problem if I set up the axis a certain way. If I slant the axis so that the acceleration is in the X direction, then the problem is easy:

F = MA

F in the x direction = MGsin(theta) = MA in the x direction
F in the y direction = N (normal force) -MGcos(theta) = 0

MA = MGsin(theta)
A = Gsin(theta)

However, I'm now asked to do the problem except with the axis the normal way, and while the free body diagram looks the same, I don't know how to precede from there because the axis are different.

.../Normal force
../
.o object
.|
.|
mg
Here is the free body diagram..

F in the Y direction: Ncos(theta) - mg ?? something like this?
F in the X direction: N sin(theta)

F = ma
Ncos(theta) - mg = MA in the y direction <- y component of acc.
Nsin(theta) = MA in the X direction <- x component of acc.
Something like that?

2. Jan 7, 2005

### marlon

Seems correct to me...

regards
marlon

3. Jan 8, 2005

### Andrew Mason

You know that the component of gravity perpendicular to the plane is balanced by the plane. Resolve the gravitational force into components that are perpendicular to the plane and along the plane (you can resolve gravity components any way you like, it is just that this is the most useful way to do it). You have to draw a line perpendicular to the plane and work out the angle between that line and the vertical (do you see that it is just theta?). What is the component of gravity along the surface? That is what provides the acceleration along the plane. Hint: It is not Nsin(theta)

AM

4. Jan 8, 2005

### marlon

yes it is...assuming that the x-axis is along the incline...the vector mg makes an angle theta with negative part of the y-axis, perpendicular to the incline. So the x-component is now the that line, opposite to theta in the triangle you use for determining the components. Thus x = mgsin(theta)

marlon

5. Jan 8, 2005

### scuddman

If the axis are normal, that is:

---> is X
|
| is Y
and the

.../Normal force
../
.o object
.|
.|
mg

free body diagram. Gravity has no X component. It is perpendicular to the X axis, and parallel to the Y axis. So gravity can't possibly have a component in the x direction. This is what is confusing to me. The normal force has both an x and y component. With the axis not tilted, the accleration must have an angle, because I didn't set the acceleration to be in the X direction, I used the axis the normal way.

So if the X component is not Nsin(theta) I don't know what else it could be.

6. Jan 8, 2005

### Andrew Mason

I agree that mgsin(theta) = ma

It doesn't matter where you put the axis. N is the normal force perpendicular to the plane so:

N= -mgcos(theta)

so Nsin(theta) is not ma

AM

7. Jan 8, 2005

### marlon

I think you are confusing two things here. The second set equations is valid when the x-axis is just horizontal and the y-axis perpendicular to it. In this case the Normal force yields N = N (sin(theta)e_x + cos(theta)e_y) and gravity yields -mge_y

If the x-axis is along the plane you get for normal force N=Ne_y and
gravity mgsin(theta)e_x - mgcos(theta)e_y

And in this case Newton's second law yields ma_y = 0 thus N = mgcos(theta)e_y. But ma_x is ofcourse not 0

In the second case (x-axis horizontal) both a_x and a_y are NOT ZERO

regards
marlon

8. Jan 8, 2005

### scuddman

It does matter where you put the axis.

That was the whole point of the exercise, I'm told that the solution has a x and y component. The point of changing the axis so that it's normal is so that the normal force isn't perpendicular to the plane of the problem.

.../Normal force
../
.o object
.|
.|
mg

-----> X
|
|
|
Y

Y Normal force (note that the normal force is slanted to the y axis)
|../
|./
|/
----------> x
|\
|..\
|....\(acc a with an X component and a y component)
mg (note that the acceleration is inbetween the x and y axis.)

Versus:
Y
..../
.../Normal force (normal force here is perpendicular to the x axis)
../ (is also parallel to the Y axis)
.o object
.|...\
.| .........\
mg...............\
X and acc a
(acc is in the x direction only, there is no y component)
in this case, yes the acc is gsin(theta) with the direction in the x direction.

in the above you also have with normal axis:

-----> X
|\
|..\
|....\
Y Acceleration, so it must contain an x and y component.

Anyways, I think that's how I'm supposed to look at the problem,

I guess long story short, it's easier to think of the normal force as perpendicular to the plane, causing you to slant the axis, but on an exam I might not think of that and try to do X and Y the inconvenient way.

9. Jan 8, 2005

### Andrew Mason

It matters only if you are going to express your answer in terms of particular unit vectors. It can't change the magnitude or direction of any of the forces. Usually one tries to choose a frame of reference that will make the problem less complicated rather than more complicated.

No. The point of the exercise was to find the magnitude and direction of the acceleration. You can express the direction in terms of any orthogonal unit vectors you want.

And I guess I don't see why that helps analyse the problem.

Why slant the axis?

You just need to resolve the forces in the way that makes it simplest to analyse the problem. Here it is to see that the force components perpendicular to the plane (N and mgcos(theta)) produce no acceleration and the component parallel to the plane (mgsin(theta)) produce acceleration. You know the direction and magnitutde so you have found its acceleration. And then you can express that in terms of an x and y component.

AM

10. Jan 8, 2005

### scuddman

Let me quote the problem:

A crate of mass M is placed on a frictionless inclined plane of angle theta.

A. Determine the acceleration of the crate after it is released.
B. It is possible to solve the problem using "standard" horizontal and vertical axes. Determine the acceleration of the crate in reference to these axes.
C. Suppose the crate is released from rest at the top of the incline, and the distance from the front edge of the crate to the bottom is "d". How long does it take the front edge to reach the bottom, and what is its speed just as it gets there?

C. I already know how to do, it's just B that was confusing. I guess it's a badly worded problem?

11. Jan 8, 2005

### dextercioby

The text seems perfectly intelligible to me.It would have helped if you had posted it from the beginning.It would have spared time and definitely it would have been no place for discussion and interpretations.
Next time,please post the extire text...

Daniel.