Release block on a moving inclined plane.

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  • #1

Homework Statement


A block of mass m is held motionless on a frictionless plane of mass M and angle of inclination θ. The plane rests on a frictionless horizontal surface. The block is released. What is the horizontal acceleration of the plane?
Physics1.png

(Introduction to Classical Mechanics by David Morin, question 3.8)

Homework Equations


F=ma

The Attempt at a Solution


Let:
N1 be the normal force exerted by the block on the plane and vice versa.
N2 be the normal force exerted by the ground on the plane and vice versa.
aM be the acceleration of the plane.
av be the vertical acceleration of the block.
ah be the horizontal acceleration of the block.

Considering the whole system:

N2 = Mg + mg

Considering the subsystem of the forces acting on the plane:

Vertical equilibrium: N2 = Mg + N1cosθ
Horizontal motion: N1sinθ = MaM

Considering the subsystem of the forces acting on the block:
Vertical motion: N1cosθ - mg = mav
Horizontal motion: N1sinθ = mah

M, m and θ are known. So I have 5 equations with 5 unknowns. However, solving the first and second equations yield: N1cosθ = mg , which, using the 4th equation will result in av = 0 and that's wrong for sure.

Looking at the answers, I see that my 3rd, 4th and 5th equations are correct, but the 1st and 2nd equations are not in the answer. So why is it that I can't use them (or why is it wrong)?
 

Answers and Replies

  • #2
Curious3141
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Never mind - I misread the problem.
 
  • #3
Curious3141
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OK, got it now.

First off, why do you think the mass m should have any net vertical component of acceleration? Momentum in the block-plane system (as observed by an inertial observer at rest with respect to the ground) has to be conserved in both the horizontal and vertical directions (since there is no net external force). To begin with, there is no horizontal or vertical momentum. After release, there is horizontal acceleration of the plane, which means there has to be horizontal acceleration of the mass as well. However, there is no vertical acceleration of the plane, which means the mass should have no vertical acceleration either.

You've correctly enumerated the forces to be considered. To simplify matters, let's call f (small f) the reaction force of the mass on the plane (and plane on mass). F is the reaction force of the ground on the plane. W is the weight of the plane (plane alone) and is equal to Mg. w is the weight of the mass, and is equal to mg. A is the acceleration of the plane. a is the acceleration of the mass. You won't even need all these parameters, as you'll see.

Consider forces acting on the mass first. You need only consider the vertical components here. Since there is zero net vertical force,

[tex]f\cos \theta = w = mg[/tex] ---eqn 1

Now consider the forces on the plane. You only need to bother with the horizontal components here. Remember that neither F nor W have any horizontal components.

[tex]f \sin \theta = MA[/tex] ---eqn 2

Divide eqn 2/eqn 1:

[tex]\tan \theta = \frac{M}{m}.\frac{a}{g}[/tex]

Therefore [itex]a = \frac{mg \tan \theta}{M}[/itex]
 
  • #4
The reason I think the vertical acceleration of the block must not be zero is because it started from rest. After releasing it, it will slide down the plane, while pushing the plane to the left. This motion down the plane will definitely have a downward acceleration component.

Looking at the solutions, it says: We cannot assume that f = mg cosθ because the plane recoils. We can see that f = mg cosθ is in fact incorrect, because in the limiting case where M = 0, we have no normal force at all.

Both your answer and mine arrives at f = mg cosθ and that's the pitfall. To be honest, I don't really understand the 'recoil' part of the solution given either.
 
  • #5
Curious3141
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The reason I think the vertical acceleration of the block must not be zero is because it started from rest. After releasing it, it will slide down the plane, while pushing the plane to the left. This motion down the plane will definitely have a downward acceleration component.

Yes, I realised this as well, but I'm uncertain how to proceed. I'll have to think about it some more.
 
  • #6
Curious3141
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Bensondros:

Do you have access to the solution? I've found an electronic copy on scribd (the link is here: http://www.scribd.com/doc/8789992/David-Morin-Introductory-Classical-Mechanics-with-Problems-and-Solutions-2004 [Broken], it's page 69/608 of the electronic version).

To get the extra equation to solve for the extra unknown, the author used the relationship between the horizontal and vertical components of the acceleration of the block and related it to tan theta (since the block slides adherently down the plane). But I have an issue with the way he did it. I wonder why he summed the horizontal components of acceleration for both the plane and the block. If displacements and velocities are being considered relative to a inertial observer at rest wrt the ground on which the plane rests, then it should simply be that [itex]a_y/a_x = -\tan \theta[/itex] shouldn't it? This does actually give [itex]f = mg\cos \theta[/itex], which he claims is wrong, so I dunno.
 
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  • #7
Thanks, I do have the solutions.

He didn't sum the horizontal components of the acceleration for both the plane and the block. He summed the horizontal distance travelled by the plane and the block.

I understand how he got his 4th equation, but I wonder why I couldn't do it my way, i.e why do my first and second equations fail when they seem perfectly logical?
 
  • #8
Curious3141
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Thanks, I do have the solutions.

He didn't sum the horizontal components of the acceleration for both the plane and the block. He summed the horizontal distance travelled by the plane and the block.

I understand how he got his 4th equation, but I wonder why I couldn't do it my way, i.e why do my first and second equations fail when they seem perfectly logical?

Sorry for the delayed reply, I've been busy.

I see now what he was doing. It's equivalent in this case whether he considers acceleration or displacement. My error was in not seeing that everything must be taken from a reference frame where the plane is stationary (and the block just slides down it). Only in this case would that ratio equal [itex]\tan \theta [/itex].

Doing this is the only way of avoiding an insufficiently determined system of equations (where there's always one more unknown than equations).

I was clearly also wrong in my first post: momentum is not conserved for this system because gravity is the net external force acting on it.

Going back to your query:

Considering the whole system:

N2 = Mg + mg ---eqn 1

Considering the subsystem of the forces acting on the plane:

Vertical equilibrium: N2 = Mg + N1cosθ ---- eqn 2
Horizontal motion: N1sinθ = MaM

Considering the subsystem of the forces acting on the block:
Vertical motion: N1cosθ - mg = mav ---eqn 3
Horizontal motion: N1sinθ = mah ---eqn 4

I've labelled your equations. You're asking why the first 2 are wrong?

Equation 2 is correct. The author also uses this equation.

Equation 1 is wrong, because you're making an implicit assumption here: that the block is in vertical equilibrium.

To see this, consider what it would mean if eqn 1 held true:

Put eqn 1 into eqn 2:

(M+m)g = Mg + N1cosθ

N1cosθ - mg = 0

Put that into equation 3:

mav = 0

which is clearly untrue, because as you yourself have observed, the block slides downward.

There is no physical way in which the block could be in equilibrium (horizontal or vertical) in the absence of friction.
 
  • #9
No need for apologies. Thanks for spotting the mistake!

And most of all, thank you so much for taking the time to respond!
 
  • #10
Curious3141
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No need for apologies. Thanks for spotting the mistake!

And most of all, thank you so much for taking the time to respond!

Not at all, glad I could help.:smile:
 

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