How Does Horizontal Acceleration Keep a Block Stationary on a Wedge?

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Homework Help Overview

The problem involves a wedge of mass M on a horizontal table with a block of mass m resting on it. The discussion centers around determining the horizontal acceleration required for the wedge to keep the block stationary relative to it, under the assumption of no friction.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the wedge's acceleration and the block's motion, questioning how the wedge's acceleration can keep the block stationary. Some participants suggest using free body diagrams to analyze forces acting on the block and the wedge.

Discussion Status

There is an ongoing exploration of the forces involved, with some participants providing equations based on their understanding of the problem. Clarifications about the nature of the forces and the role of inertia are being discussed, indicating a productive dialogue on the concepts involved.

Contextual Notes

Participants note the importance of understanding the relative motion between the block and the wedge, as well as the implications of frictionless conditions. There is also mention of the need for sufficient horizontal force for the block to keep up with the wedge's acceleration.

Rococo
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Homework Statement



A wedge of mass M rests on a horizontal table. A block of mass m rests on the wedge.

http://imgur.com/cS02MvM

(a) What horizontal acceleration, a, must M have relative to the table to keep the block stationary relative to the wedge, assuming no friction?

(b) What horizontal force F must be applied to the system to achieve this result, assuming a frictionless table top?

Homework Equations



F=(m+M)a

The Attempt at a Solution



http://imgur.com/etif1qn

All I have so far is this diagram as I don't understand the concept involved. Its not clear to me why the horizontal acceleration of the wedge could keep the block stationary. Also, I'm not sure whether this acceleration would be to the left or to the right.

Also I have read through other similar threads but still can not understand the problem.
 
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Rococo said:

Homework Statement



A wedge of mass M rests on a horizontal table. A block of mass m rests on the wedge.

cS02MvM.jpg


(a) What horizontal acceleration, a, must M have relative to the table to keep the block stationary relative to the wedge, assuming no friction?

(b) What horizontal force F must be applied to the system to achieve this result, assuming a frictionless table top?

Homework Equations



F=(m+M)a


The Attempt at a Solution



etif1qn.jpg


All I have so far is this diagram as I don't understand the concept involved. Its not clear to me why the horizontal acceleration of the wedge could keep the block stationary. Also, I'm not sure whether this acceleration would be to the left or to the right.

Also I have read through other similar threads but still can not understand the problem.

Start by writing the FBD equations. Note that for mass m to be stationary with respect to the wedge, its vertical acceleration must be zero, and it horizontal acceleration must be equal to that of M.
 
Rococo said:
Its not clear to me why the horizontal acceleration of the wedge could keep the block stationary.
It does not keep the block stationary. It keeps it stationary relative to the wedge.
To get the concept, imagine you have a block of ice on a flat tray. If you pull the tray swiftly to one side the ice block will be left behind. Now make the tray slightly wedge shaped so that the ice has to rise slightly in order to slide off. It will still slide off if you jerk the tray fast enough. So increase the slope a bit more...
 
Start by writing Free body diagrams.Create equations applying
ƩFx=max &
ƩFy=may

This will give all the needed equations.
 
voko said:
Start by writing the FBD equations. Note that for mass m to be stationary with respect to the wedge, its vertical acceleration must be zero, and it horizontal acceleration must be equal to that of M.

Here's what I got:

Ncos30° = mg

Nsin30° = F = (m+M)a

Is this right?
 
haruspex said:
It does not keep the block stationary. It keeps it stationary relative to the wedge.
To get the concept, imagine you have a block of ice on a flat tray. If you pull the tray swiftly to one side the ice block will be left behind. Now make the tray slightly wedge shaped so that the ice has to rise slightly in order to slide off. It will still slide off if you jerk the tray fast enough. So increase the slope a bit more...

What's the reason this happens? Is it just as a result of the inertia of the block?
 
Rococo said:
Here's what I got:

Ncos30° = mg

Nsin30° = F = (m+M)a

Is this right?
The first equation, yes. But is F above the force applied to the wedge, as in the question? Why would that be equal to N sin 30? What body did you study an FBD of to obtain it?
 
Rococo said:
What's the reason this happens? Is it just as a result of the inertia of the block?
Yes. For the block to accelerate as fast as the wedge it needs a sufficient horizontal force. If the slope is too low or the friction inadequate then it won't keep up.
 
haruspex said:
The first equation, yes. But is F above the force applied to the wedge, as in the question? Why would that be equal to N sin 30? What body did you study an FBD of to obtain it?

Yes, F is the force applied to the wedge.

I realize F is not equal to Nsin30° now. The horizontal force acting on the block is Nsin30°. The horiztonal force acting on the wedge is F.

As voko said, the horizontal acceleration of the block must be equal to the horizontal acceleration of the wedge. So, since acceleration is force divided by mass I get this:

(Nsin30°)/m = F/(m+M)
 
  • #10
Rococo said:
Yes, F is the force applied to the wedge.

I realize F is not equal to Nsin30° now. The horizontal force acting on the block is Nsin30°. The horiztonal force acting on the wedge is F.

As voko said, the horizontal acceleration of the block must be equal to the horizontal acceleration of the wedge. So, since acceleration is force divided by mass I get this:

(Nsin30°)/m = F/(m+M)
Yes.
 
  • #11
haruspex said:
Yes.

Using the two equations:

(Nsin30°)/m = F/(m+M)
Ncos30° = mg

I was able to rearrange to get the following:

F/(m+M) = gtan30°
a = gtan30°
F = (m+M)gtan30°
 
  • #12
Rococo said:
Using the two equations:

(Nsin30°)/m = F/(m+M)
Ncos30° = mg

I was able to rearrange to get the following:

F/(m+M) = gtan30°
a = gtan30°
F = (m+M)gtan30°
Looks good.
 

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