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(somewhat complex inclined plane problem) why is this wrong?

  • #1
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1.png

Note:- All surfaces given here are frictionless.
To find:-
Acceleration a

Relevant eqs:-
F = ma

Attempt at solution:-
The equations I've gotten so far:-
1) N1sin(theta) = m2a
2) m1gsin(theta) = m1a1
3) m1gcos(theta) - N1 = m1a2

So far, 3 eqs, 4 unknowns.

For the 4th eq, I did a2sin(theta) = a i.e the horizontal component of a2 = a.
However, this is wrong. The correct last eq is a2 = asin(theta) i.e. the component of a perpendicular to the plane = a2. Why is my answer wrong? I don't see the difference in relating them in these different ways.

Note:- I've verified for a fact that my answer is incorrect(a2sin(theta) = a)
 
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Answers and Replies

  • #2
gneill
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Can you add a problem statement that specifies what it is that's to be calculated?
 
  • #3
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Can you add a problem statement that specifies what it is that's to be calculated?
done
 
  • #4
PeroK
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done
Try using conservation of energy and momentum. As both the wedge and the block are accelerating, it's difficult to use the forces between them.
 
  • #5
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Try using conservation of energy and momentum. As both the wedge and the block are accelerating, it's difficult to use the forces between them.
Tbh, this post wasn't intended to ask for ways to solve the above problem. I've already come up with 3 eqs but the 4th one I've come up with is wrong. I'm trying to understand where I went wrong and why the correct answer is...well, correct.
 
  • #6
PeroK
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Tbh, this post wasn't intended to ask for ways to solve the above problem. I've already come up with 3 eqs but the 4th one I've come up with is wrong. I'm trying to understand where I went wrong and why the correct answer is...well, correct.
Analyse the motion from the reference frame of the wedge.
 
  • #7
haruspex
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I did a2sin(theta) = a
The actual acceleration of the wedge is a. It has a component a2 in the direction of a2 (or the surfaces would not stay touching) and a component normal to a2.
If you try to work the other way around, yes a2 can be thought of as contributing a2 sinθ to a, but it is not the only contributor.
 
  • #8
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The actual acceleration of the wedge is a. It has a component a2 in the direction of a2 (or the surfaces would not stay touching) and a component normal to a2.
If you try to work the other way around, yes a2 can be thought of as contributing a2 sinθ to a, but it is not the only contributor.
What are the other contributors?
 
  • #9
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Analyse the motion from the reference frame of the wedge.
I've already solved the problem using that method. I'm just trying to solve it from an inertial reference frame too.
 
  • #10
haruspex
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What are the other contributors?
a has a component normal to a2.
The point is that the relationship between a and a2 is determined by the plane of conact between the mass and the wedge. For the mass and the wedge to remain in the same relationship in that plane the two accelerations must have the same component normal to it. a2 is entirely normal to that plane, whereas only a component of a is normal to it.
 
  • #11
PeroK
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I've already solved the problem using that method. I'm just trying to solve it from an inertial reference frame too.
The point is that ##a_2' = 0##, where ##a_2'## is the normal acceleration of the block in the wedge's frame. Hence, in the inertial frame ##a_2 = A_2## where ##A_2## is the acceleration of the wedge in the normal direction. Hence ##a_2 = asin \theta## (by resolving ##a## into normal and tangential components).
 
  • #12
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a has a component normal to a2.
The point is that the relationship between a and a2 is determined by the plane of conact between the mass and the wedge. For the mass and the wedge to remain in the same relationship in that plane the two accelerations must have the same component normal to it. a2 is entirely normal to that plane, whereas only a component of a is normal to it.
The horizontal component of the normal force due to the block is the net force causing wedge to accelerate towards the right so why is saying that the horizontal component of the perpendicular acceleration = the acceleration of the wedge? Don't they both HAVE to be equal if they're moving together?
 
  • #13
haruspex
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The horizontal component of the normal force due to the block is the net force causing wedge to accelerate towards the right so why is saying that the horizontal component of the perpendicular acceleration = the acceleration of the wedge? Don't they both HAVE to be equal if they're moving together?
We were discussing accelerations, not forces.
Yes, the horizontal component of the normal force corresponds to the acceleration a, but what is its relationship to acceleration a2?
 
  • #14
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We were discussing accelerations, not forces.
Yes, the horizontal component of the normal force corresponds to the acceleration a, but what is its relationship to acceleration a2?
Well, that's what I said.
The horizontal component of the normal force due to the block is the net force causing wedge to accelerate towards the right so why is saying that the horizontal component of the perpendicular acceleration = the acceleration of the wedge wrong? Don't they both HAVE to be equal if they're moving together?
 
  • #15
haruspex
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why is saying that the horizontal component of the perpendicular acceleration = the acceleration of the wedge? Don't they both HAVE to be equal if they're moving together
They are not moving together in the horizontal direction. They are moving together in the direction perpendicular to the plane of the wedge.
 
  • #16
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They are not moving together in the horizontal direction. They are moving together in the direction perpendicular to the plane of the wedge.
Wait, what? How are they not moving together in the horizontal? If they were not, wouldn't the block leave the surface of the wedge?
 
  • #17
haruspex
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Wait, what? How are they not moving together in the horizontal? If they were not, wouldn't the block leave the surface of the wedge?
Suppose you mark on the wedge the initial position of the block. After the system has been moving a little while, where is the block in relation to the mark? Have both moved the same horizontal distance?
 
  • #18
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Suppose you mark on the wedge the initial position of the block. After the system has been moving a little while, where is the block in relation to the mark? Have both moved the same horizontal distance?
Oh yeah. The block's net acceleration will actually be towards the left. That clears things up, thanks!
 

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