# Finding the acceleration of a block of ice on a ramp

## Homework Statement

A a block of ice with mass mA rests on a wedge of ice with mass mB=4.33 kg that is cut so that the surface of the wedge makes an angle of θ=38.1° with the ground. There is no friction between the pieces of ice or between the ice and the ground. At time t=0, the blocks are released from rest in the configuration shown. What is the acceleration of mA in the direction parallel to the ground (the x-direction indicated in the figure)? Express your answer in m/s2. No, you don't need to know the value of mA. F=ma
F=mg

## The Attempt at a Solution

∑FxB= -NSinθ = mba[/B]

#### Attachments

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Sorry but can't seem to be able to view the image you've uploaded.

gneill
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Jphysics,

Your image link goes to a site that is behind a login barrier. Please use the UPLOAD button (Bottom right of the edit panel) to upload a local copy your image file. We prefer to have local copies of images or other attachments as we find that off-site links or materials tend to 'evaporate' over time.

neilparker62
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I would guess you need to calculate the horizontal acceleration of Mass a if it were on a stationary wedge. Then calculate the acceleration of B (based on equal but opposite horizontal force) and add the two. But perhaps I need to see that diagram first!

## Homework Statement

A a block of ice with mass mA rests on a wedge of ice with mass mB=4.33 kg that is cut so that the surface of the wedge makes an angle of θ=38.1° with the ground. There is no friction between the pieces of ice or between the ice and the ground. At time t=0, the blocks are released from rest in the configuration shown. What is the acceleration of mA in the direction parallel to the ground (the x-direction indicated in the figure)? Express your answer in m/s2. No, you don't need to know the value of mA. F=ma
F=mg

## The Attempt at a Solution

∑FxB= -NSinθ = mba[/B]
Jphysics,

Your image link goes to a site that is behind a login barrier. Please use the UPLOAD button (Bottom right of the edit panel) to upload a local copy your image file. We prefer to have local copies of images or other attachments as we find that off-site links or materials tend to 'evaporate' over time.

I am just confused as this question uses a different coordinate system than I am used to:
I worked out the answer to be -7.69m/s2
but I am not sure if the negative acceleration makes sense

I would guess you need to calculate the horizontal acceleration of Mass a if it were on a stationary wedge. Then calculate the acceleration of B (based on equal but opposite horizontal force) and add the two. But perhaps I need to see that diagram first!

This is the link to my solution set so far. I found it hard to communicate without showing my freebody diagram: https://drive.google.com/open?id=1rwSUy2l8RXOEdg4HPofT-AkZ22r_U_jP

gneill
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I've inserted a local copy of your image in your Original Post. We find that links to, and the contents of off-site materials often vanish over time, which can destroy the value of a thread.

stockzahn
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Which direction you suppose the mass ##m_A## will move?

You didn't take into account all forces in the ##y##-direction of the mass ##m_B## (wedge) - there is only the weight of the wedge and the reaction force of the floor. What about the normal force of the mass ##m_A## acting on the wedge?

Which direction you suppose the mass ##m_A## will move?

You didn't take into account all forces in the ##y##-direction of the mass ##m_B## (wedge) - there is only the weight of the wedge and the reaction force of the floor. What about the normal force of the mass ##m_A## acting on the wedge?
I think the massA would move rightwards.

Actually, I have included the normal force and it is denoted N on my diagram.

stockzahn
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I think the massA would move rightwards.

I think that's good intuition. That problem looks easier than it is, only applying Newton's 2nd law will not be sufficient to solve it.

neilparker62
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Typically in inclined plane questions such as this mg (of block resting on the plane) resolves into components mgcosθ perpendicular to the inclined plane and mgsinθ parallel to the plane. So your normal force should just be mgcosθ.

stockzahn
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Actually, I have included the normal force and it is denoted N on my diagram.

In the the schematic, yes (although there seems to be a mistake as far as I understand it). But the equation should read

$$\sum F_y = 0 = F_{Floor}-m_B\;g-N\;cos\theta$$

I think the massA would move rightwards.

Actually, I have included the normal force and it is denoted N on my diagram.
So
Typically in inclined plane questions such as this mg (of block resting on the plane) resolves into components mgcosθ perpendicular to the inclined plane and mgsinθ parallel to the plane. So your normal force should just be mgcosθ.

How do I solve for that when I am not given mass of block A. Also, based on the diagram that I have drawn the Normal force would be mg/Cosθ

haruspex
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So your normal force should just be mgcosθ.
I assume you are referring to the normal force between block and wedge.
That would mean the block has no acceleration normal to the slope of the wedge, but since the wedge moves that will not be the case.
based on the diagram that I have drawn the Normal force would be mg/Cosθ
It will not be that either.

neilparker62
Homework Helper
I assume you are referring to the normal force between block and wedge.
That would mean the block has no acceleration normal to the slope of the wedge, but since the wedge moves that will not be the case.
Quite right - sorry for that. But the point I wanted to make was that mg resolves into components parallel and perpendicular to the inclined plane - hence mgcosθ and not mg/cosθ. On a fixed plane the perpendicular component will be equal and opposite to the normal force whereas in this problem it presumably causes acceleration of the combined mass to the left.

I can't quite see how mass A will not be needed since if we consider an extreme case with Mass A << Mass B, the horizontal acceleration of Mass A will reach a limiting value of gsinθcosθ. Seems to be a somewhat complicated problem if this ref is relevant (?)

http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node80.html

haruspex
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I can't quite see how mass A will not be needed
Quite. It is, as you note, demonstrably false.

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haruspex
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This is the link to my solution set so far. I found it hard to communicate without showing my freebody diagram: https://drive.google.com/open?id=1rwSUy2l8RXOEdg4HPofT-AkZ22r_U_jP
You have been inconsistent with the sign of the acceleration a. In one place you have N sin(θ)=mAa, in another it is -mAa.
As others have noted, finding the normal force is nontrivial. Instead, you need an equation relating the three accelerations and representing the fact that the block remains on the surface of the wedge.

You have been inconsistent with the sign of the acceleration a. In one place you have N sin(θ)=mAa, in another it is -mAa.
As others have noted, finding the normal force is nontrivial. Instead, you need an equation relating the three accelerations and representing the fact that the block remains on the surface of the wedge.
So would normal force be N=mgCosθ?

haruspex
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So would normal force be N=mgCosθ?
I showed that was wrong in post #15.
Stop trying to guess the normal force - you will not succeed - and follow the line I described at the end of post #18.

First, I would mark down the directions of the accelerations of the two objects along with the force diagram. As you said earlier, the ice block would move horizontally(with respect to the floor) so you can mark that off with an acceleration value quite easily(which would be an unknown in your equations). The acceleration of B is not so straightforward. It moves on the wedge, but this motion is relative to the wedge. So if you assume B moves on the surface of A with some acceleration, then you need to use relative acceleration equations to figure out it's acceleration relative to the floor. It is this acceleration value that you should use in the force equations. You've used ##a## as the unknown variable for acceleration in the solution you posted. In reality you need two different variables to account for the two accelerations of the two objects A and B. But you would already know the directions of these accelerations.

Next, you have the two reaction forces between A and B and between A and the floor. These two forces are again two variables you don't know. This runs up the tally of total unknowns in the system to 4. But you do have 4 equations you can write down(as you you already have). From there on the problem is solving 4 simultaneous equations for 4 unknowns. You actually only need to solve for the two accelerations.

As was stated earlier, you are not asked for the normal forces in the question, so just solving for the accelerations is really sufficient. But you can calculate the forces quite easily once you know the accelerations. While the reaction force between A and B would ##m_BgCos(\theta)## if A were forced to be stationary, since A is in motion, it would not be equal to that value in general.

But what about ##m_B##? Is that not an unknown as well? In some physics problems, certain variables just get cancelled out because of the structure of the problem. This is such a case where you will find ##m_B## disappear.

haruspex
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This is such a case where you will find ##m_B## disappear.
Have you confirmed that or are you just trusting the information in post #1?
As has been shown in an earlier post, consideration of extreme values indicates the mass of B is needed. Also, by dimensional analysis, you can prove that if the mass of B is irrelevant then so must be the mass of A.

neilparker62
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In the the schematic, yes (although there seems to be a mistake as far as I understand it). But the equation should read

$$\sum F_y = 0 = F_{Floor}-m_B\;g-N\;cos\theta$$
Just a conceptual question here - if (say) we imagine the frictionless floor to be the 'pan' of a (wide diameter!) weighing scale, are we saying that the weight of the moving system is less than (Ma + Mb)g ?

Merlin3189
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... are we saying that the weight of the moving system is less than (Ma + Mb)g ?
I'd have thought so. Why not?
If I stood on a weighing machine holding a mass in my hand. When I release it so that it falls, between the moment it leaves my hand until the moment it lands on my toe, the reading on the machine will be less. Sliding down a slope is just a less extreme case.

• neilparker62
neilparker62
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Aha - Eureka moment for me - albeit fairly obvious when you put it that way. So when considering horizontal motion of block B with block A sliding down, the 'effective mass' we use in equation F=ma will be less than Ma+Mb ?