Two boxes sliding down inclined plane

[zeralda21]

Homework Statement

A block with m=20kg is sliding on top of a block of M=10 kg that is sliding down an inclined plane with 30 degrees to the horizontal. All surfaces are frictionless. Calculate the acceleration of the blocks.

Homework Equations

Newtons second law: F⃗ net=ΣF⃗ =ma⃗

The Attempt at a Solution

For sake of generality, let's say that the angle is indefinite.
I have sketched a free-body diagram and F⃗ net=-mgsin(\theta)-Mgsin(\theta)=(m+M)a⃗ since the boxes will moves as one unit. Solving for acceleration gives
a⃗=-gsin(\theta) which is wrong. I think it is something with the signs. They both have the same sign since they move in the same direction(downward=negative) but even if both positive would yield wrong answer.

 

[Doc Al]

Why do you think your answer is wrong? (They probably just want the magnitude of the acceleration, not the direction.)

 

[BruceW]

it is correct. But think about what you have calculated. The total force on both the blocks, so this will give you the acceleration of what? (Hint: it is not the acceleration of one or other of the blocks).

 

[zeralda21]

Yes, magnitude is of importance here. In this case, with an angle 30 degrees, the acceleration would be \frac{g}{2} which still is a wrong solution. The equation is supposed to be

(m+M)a=(m-M)gsin(\theta) which in the end yields \frac{1}{6}g but I fail to see why the forces are in the opposite direction of each other.

 

[Doc Al]

I guess I don't understand the problem. Is the incline fixed in place?

If this is from a textbook, can you give the name and problem number.

 

[zeralda21]

I'll provide a figure in just a second. No, this one appeared in an old test.

http://tinypic.com/view.php?pic=9zpwec&s=6

 

[BruceW]

haha, the pulley makes a big difference :) Also, I think your answer before would have been right (as Doc Al said), but the inclusion of the pulley makes it a bit more complicated.

 

[Doc Al]

Now we have a completely different problem (from the one I was envisioning).

Start by drawing free body diagrams for each block.

 

[zeralda21]

Yes, that was a mistake of mine. Alright, the free body diagrams. Both are being pulled with a Force from the pulley, tension, call it F_{T} There is also a gravitational force in the -y direction with magnitude mg. This can(for both blocks) be written mgsin(\theta) which is parallel to the incline.
Also, for both blocks, there is a normal force, perpendicular to the incline.

So both free-body diagrams include the same forces, but with different magnitudes.

 

[Doc Al]

Yes, that was a mistake of mine. Alright, the free body diagrams. Both are being pulled with a Force from the pulley, tension, call it F_{T}

Good.

There is also a gravitational force in the -y direction with magnitude mg. This can(for both blocks) be written mgsin(\theta) which is parallel to the incline.

Good. Of course the blocks have different masses. (Personally, I would choose the positive direction to be down the incline, but it doesn't matter as long as you are consistent.)

So both free-body diagrams include the same forces, but with different magnitudes.

OK. Now apply Newton's 2nd law to each. (Careful with signs.)

 

[zeralda21]

Alright. I actually not sure how to include the normal force. Is it F_{N}sin(120)? For block 1: F_{net}=ma=-mgsin(30)+F_{T}+F_{N}sin(120)
For block 2: F_{net}=ma=-mgsin(30)+F_{T}+F_{N}sin(120)

 

[Doc Al]

Alright. I actually not sure how to include the normal force. Is it F_{N}sin(120)? For block 1: F_{net}=ma=-mgsin(30)+F_{T}+F_{N}sin(120)
For block 2: F_{net}=ma=-mgsin(30)+F_{T}+F_{N}sin(120)

The normal force acts perpendicularly to the direction of motion. Since there's no friction, you don't need to worry about it. And you would never add perpendicular force components!

Be sure to use different symbols for the masses. What's the relationship between the acceleration of each block?

 

[zeralda21]

Yes of course, i should have known that. If it is perpendicular, angle is 90 degrees and sin(90)=0. We label the mass of block 1, m, and mass of block 2 by M.

So for block 1:

F_{net}=ma=-mgsin(30)+F_{T} which yields a=-\frac{g}{2}+\frac{F_{T}}{m}.

For block 2:

F_{net}=Ma=-Mgsin(30)+F_{T} which yields a=-\frac{g}{2}+\frac{F_{T}}{M}.

So I will have to find F_{T}? Is that possible even if the system is not in equilibrium?

 

[Doc Al]

Careful with the sign of the acceleration.

Let "a" stand for the magnitude of the acceleration. Pick one of the boxes (say M) and assume it slides down the incline. What would be its acceleration? What would be the acceleration of the other box?

As far as the tension is concerned, you have two equations and two unknowns (tension and acceleration). So there shouldn't be a problem in solving for both.

 

[zeralda21]

I fail too see why they should'nt have the same sign. If the lower block is moving with an acceleration a, and the other block moves exactly the same on top of it. Is not the acceleration the same for this block too? At least that is what my intuition tells me. Maybe I could think that one of them is 0 but different sign? I can't see that actually..

 

[mfb]

They cannot both move downwards, the string across the pulley has a fixed length. If one block goes down, the other one has to go up - and the question is, how are the two accelerations related (via the fixed string length).

 

[zeralda21]

They cannot both move downwards, the string across the pulley has a fixed length. If one block goes down, the other one has to go up - and the question is, how are the two accelerations related (via the fixed string length).

Of course. Oh well, that explains it. Thanks Doc Al and mrb.