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Pedagogical machine acceleration when force is 0

  • Thread starter PhizKid
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475
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1. Homework Statement
V5MO7.png



2. Homework Equations
F = ma

3. The Attempt at a Solution
I'm completely unsure as to what the horizontal reaction force on m1 actually is. Any horizontal reaction force coming from m3 being constrained to be in between the two sides of m1 as shown will not cause m1 to move which is clear from the diagram. There is also no friction between m1 and m2 therefore I have no idea what reaction force is actually causing m1 to move. Thanks!
 
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There is a force at the wheel. A force from M3 is relevant, too (it slows acceleration of M1 a bit).
 
475
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There is a force at the wheel. A force from M3 is relevant, too (it slows acceleration of M1 a bit).
What force from m3? m3 doesn't come into contact with m1 until after m1 has accelerated backwards a bit.

Also I'm not seeing how there can be some form of a hinge force on the wheel because the string rolling over the frictionless wheel is massless

Thanks
 
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haruspex

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What force from m3? m3 doesn't come into contact with m1 until after m1 has accelerated backwards a bit.
I think you have to assume that m3 fits snugly into the shaft in m1. At least see if it gives the desired answer.
Also I'm not seeing how there can be some form of a hinge force on the wheel because the string rolling over the frictionless wheel is massless
There will be tension in the string. It acts leftwards and downwards on the pulley, so must be balanced by a force from m1 acting upwards and to the right.
 
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I think you have to assume that m3 fits snugly into the shaft in m1. At least see if it gives the desired answer.

There will be tension in the string. It acts leftwards and downwards on the pulley, so must be balanced by a force from m1 acting upwards and to the right.
If we assume m3 fits exactly between the shaft then why do we even care about the contact forces there? m3 will exert a contact force to the left and to the right on m1 which will cancel out on m1 as a whole so it seems the reaction force of interest is the one caused by the weight of m3 pressing down on the wheel but I'm not sure what u mean by 'the tension acts leftwards and downwards on the pulley' because tension is along the string. Thanks!
 

haruspex

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If we assume m3 fits exactly between the shaft then why do we even care about the contact forces there? m3 will exert a contact force to the left and to the right on m1 which will cancel out on m1 as a whole
Certainly you can treat M1+M3 together for horizontal acceleration in an appropriate equation. What would that equation be?
so it seems the reaction force of interest is the one caused by the weight of m3 pressing down on the wheel but I'm not sure what u mean by 'the tension acts leftwards and downwards on the pulley' because tension is along the string. Thanks!
Think of the bit of string going around the pulley as a 'free body'. It is subject to a horizontal and a vertical force from the tension. Since it has no mass, this must be exactly cancelled by a force from the pulley (i.e. from M1). What would the magnitude and direction of that force be?
 
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Certainly you can treat M1+M3 together for horizontal acceleration in an appropriate equation. What would that equation be?

Think of the bit of string going around the pulley as a 'free body'. It is subject to a horizontal and a vertical force from the tension. Since it has no mass, this must be exactly cancelled by a force from the pulley (i.e. from M1). What would the magnitude and direction of that force be?
If we are going to treat m1 and m3 as moving as one body for this situation then shouldn't we include m2 as well in the horizontal acceleration? As for the reaction force due to the tension in the string on the wheel, wouldn't it be magnitude T (tension) and directed 45 degrees
 

haruspex

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If we are going to treat m1 and m3 as moving as one body for this situation
only in the horizontal direction. In the vertical you can treat M2 and M1 as one body for the normal force from the floor, but M3's contribution will be less because of its downward acceleration.
then shouldn't we include m2 as well in the horizontal acceleration?
No, because it has horizontal acceleration relative to the other two.
As for the reaction force due to the tension in the string on the wheel, wouldn't it be magnitude T (tension) and directed 45 degrees
Right direction, wrong magnitude.
 
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only in the horizontal direction. In the vertical you can treat M2 and M1 as one body for the normal force from the floor, but M3's contribution will be less because of its downward acceleration.

No, because it has horizontal acceleration relative to the other two.

Right direction, wrong magnitude.
I don't understand where we are using m3 the way you are stating. There is no net contribution of horizontal force on m1 due to m3 if we assume it fits exactly into the shaft so wouldn't we just have, in the frame of m1, 0 = -sqrt2T - (m1 + m3)a1, m3a3_y = T - m3g, m2a'2 = T - (m1 + m3)a1? Also I'm not sure where your mention of the normal force comes into play here.
 

haruspex

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I don't understand where we are using m3 the way you are stating. There is no net contribution of horizontal force on m1 due to m3 if we assume it fits exactly into the shaft so wouldn't we just have, in the frame of m1, 0 = -sqrt2T - (m1 + m3)a1,
That is using M3 the way I meant.
m3a3_y = T - m3g, m2a'2 = T - (m1 + m3)a1?
I don't understand where that last equation comes from. What does M2 care about M1+M3? Also need an equation relating accelerations (representing the fact that the string is constant length).
Also I'm not sure where your mention of the normal force comes into play here.
Since there's no friction (right?) you don't care about the normal force. I only mentioned it for completeness.
 
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So are we treating m1 and m3 as moving together then horizontally because m3 is assumed to be stuck totally inside the shaft but not m2 because it is moving relative to m1 unlike the case where m2 doesn't slip on m1 like in the previous problem so in that case we would consider m2 and m1 and m3 to be moving all together horizontally. That all fine? Also, should it be sqrt2*T or T in the equation involving m1,m3 because I think sqrt2*T is the magnitude of the resultant force acting against the tension. Thanks!
 

haruspex

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So are we treating m1 and m3 as moving together then horizontally because m3 is assumed to be stuck totally inside the shaft but not m2 because it is moving relative to m1
Yes.
should it be sqrt2*T or T in the equation involving m1,m3 because I think sqrt2*T is the magnitude of the resultant force acting against the tension. Thanks!
Yes, it's T in each of the two directions.
 

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