# Block on a moving incline

1. Feb 20, 2010

### AirForceOne

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

A small block of mass m rests on the sloping side of a triangular block of mass M which itself rests on a horizontal table. Assuming all surfaces are frictionless, determine the magnitude of the force F that must be applid to M so that m remains in a fixed position relative to M (that is, m doesn't move on the incline).

2. Relevant equations

F = ma

3. The attempt at a solution

I find the forces on block m due to gravity and the force F. So the net force on that block is:
Fcos(theta)+mgsin(theta)=ma

Since the block m isn't moving, I set the net force on it equal to zero (?):
Fcos(theta)+mgsin(theta)=0

And I solve for F:
F= -mgtan(theta)

F=(m+M)gtan(theta)

What am I missing? I've been scratching my head for awhile now...

2. Feb 20, 2010

### dr_k

Draw a FBD for the small block, and prove that its horizontal acceleration must be equal to
$\rm a_x = g tan\theta$, if its vertical acceleration is to equal zero, i.e. $\rm a_y =0$. Please note, there is no F in the FBD diagram for the small block, only the externally applied N and mg. You could now do a FBD for the large block, making sure you include the action-reaction pair, to determine the necessary F. But a quicker way would be to isolate both blocks together and apply $\rm F = (m+M) a_x$, since the two move as one.

So, step one, draw a FBD for the small block and figure out how the contact force N breaks up into vertical and horizontal components. Let me know how it's going, and we'll see if you need some more hints.

Last edited: Feb 20, 2010
3. Feb 20, 2010

### AirForceOne

Isn't the normal force perpendicular to the surface of the incline? Why would there be x and y components of it.

Thanks

4. Feb 20, 2010

### dr_k

http://img32.imageshack.us/img32/8211/blocks1d.jpg [Broken]

Last edited by a moderator: May 4, 2017
5. Feb 20, 2010

### AirForceOne

I gotcha.

x-dir: Nsin(theta)=ma
y-dir: Ncos(theta)-mg=0

a=gtan(theta)

How come the net forces will not be the same if I decide to split mg into x and y components instead of splitting the normal force? How do I know which to split?

6. Feb 20, 2010

### dr_k

Good job!

If you look at the picture, the system is going to accelerate in the horizontal direction (+x for my picture). Therefore the net force on the system is a vector pointing in, for my picture, the +x direction as well. You can make a coordinate system anywhere you want, but then you would have to break the applied F up into components as well. The coordinate system I chose makes solving the problem as simple as possible. For example, consider solving circular motion problems with x-y cartesian coordinates, you can do it, but it just makes the problem more difficult, as opposed to using r-$\rm \theta$ unit vectors (cylindrical coordinates) which vastly simplifies things.

Now that you've got a handle on this problem, try to see if you can get the required result by isolating the large mass M, and using Newton's 2nd postulate, $\rm {\bf F}_{net} = M {\bf a}$. Here's the picture:

http://img69.imageshack.us/img69/894/blocks2.jpg [Broken]

Note: (1) Newton's 3rd postulate: the force N must be applied to the mass M, equal magnitude and opposite direction, since M applied N on little m. This is the action-reaction pair I mentioned earlier. (2) I moved the contact force $\rm \eta$, the contact force from the floor, a little to the left so you could see where Mg and $\rm \eta$ are being applied.

Last edited by a moderator: May 4, 2017
7. Feb 20, 2010

### AirForceOne

Okay I think I understand now...pick a coordinate system and stick with it...

For the larger block:
x-dir: F-Nsin(theta)=M(gtan(theta))
y-dir: Ncos(theta)-mg=0

Plugging N=mg/cos(theta) into the x-direction net force:
F-mgtan(theta)=Mg(tan(theta))
F=Mg(tan(theta))+mgtan(theta)

F=(M+m)gtan(theta)

Great! Thanks a lot!

8. Feb 20, 2010

### dr_k

$\rm \eta-Mg-NCos\theta = 0$ (this is from the FBD for M, but you don't need it)

Yes, good job! You needed information from doing two FBDs, for (1) m and M, or (2) m and (m+M) as a system. When you do (m+M) as a system, the action-reaction pair doesn't show up, it's "internal" to the system -- they cancel by Newton's 3rd postulate.

Now, since you're becoming a physicist, you should look at your result for $\rm a_x = g tan \theta$, and ask if it makes physical sense when $\rm \theta = 0^o$, and $\rm \theta = 90^o$.

9. Feb 20, 2010

### AirForceOne

Wait so how do you find the normal force of the larger block? Haha.

Thanks!

10. Feb 20, 2010

### AirForceOne

That was a serious question, by the way.

11. Feb 20, 2010

### dr_k

Oh? It was? lol.

I think I wrote it down earlier. Up there somewhere...hang on, there it is:

https://www.physicsforums.com/latex_images/25/2590137-0.png [Broken]

from the summation $\rm F_{y_{net}} = M a_y$

Last edited by a moderator: May 4, 2017
12. Feb 21, 2010

### AirForceOne

I meant: What is the magnitude of lower-case n, the normal force on the larger block?

Because now if I plug that equation into the net force equation for the x direction, I still have an unknown variable n...

EDIT: Nevermind. I don't need to use that equation (N-Mg-Ncos(theta)=0) because that's not what I want when finding F anyways...so that's what you mean by "you don't need it" haha...my bad

Last edited: Feb 21, 2010
13. Feb 21, 2010

### ideasrule

Look at the equation n-Mg-Ncos(theta)=0. You don't know N, so look at the other equations you wrote down to find it.