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Pedagogical Machine

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1. Homework Statement

A “pedagogical machine” is illustrated in the sketch (attached image). All surfaces are frictionless. What force ##F## must be applied to ##M_1## to keep ##M_3## from rising or falling?

2. Homework Equations

$$\sum_{}^{} \vec{F} = m \vec{a}$$

3. The Attempt at a Solution

I have chosen a coordinate system such that all particles have positive coordinates. My ##x##-axis points to the right and my ##y##-axis points upwards. The length of the rope remains constant, and it can be expressed as:
$$l = x_p - x_2 + \frac{\pi R}{2} + y_p - y_3$$
The pulley is located at ##(x_p,y_p)##.
Differentiating twice with respect to time and rearranging:
$$\ddot{x}_2 + \ddot{y}_3 = \ddot{x}_p$$
Since the pulley and ##M_1## are fixed:
$$\ddot{x}_2 + \ddot{y}_3 = \ddot{x}_1$$
Since ##\dot{y}_3 = 0## for all ##t##, ##\ddot{y}_3 = 0##.
In other words:
$$\ddot{x}_1 = \ddot{x}_2$$
Now, applying Newton's second law to each mass:
$$T - M_3 g = 0 ⇒ T = M_3 g$$
$$T = M_2 \ddot{x}_2$$
$$F - T = M_1 \ddot{x}_1$$
Where ##T## is the tension in the rope.
$$M_3 g = M_2 \ddot{x}_2 ⇒ \ddot{x}_2 = \frac{M_3 g}{M_2} = \ddot{x}_1$$
$$F = T + M_1 \ddot{x}_1 = M_3 g (1 + \frac{M_1}{M_2})$$
$$F = M_3 g \frac{M_1 + M_2}{M_2}$$
Which doesn't match the clue given in my book. According to my book, if all three masses are equal, then ##F = 3Mg##.
Where did I go wrong?
 

Attachments

haruspex

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How do you get your equation ##F-T=M_1\ddot x_1##? What are the forces acting on M1?
 
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How do you get your equation ##F-T=M_1\ddot x_1##? What are the forces acting on M1?
In the ##x##-direction, there's ##F## and ##T##.
##T## acts on the pulley, which is a part of ##M_1##.
All surfaces are frictionless.
 

haruspex

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In the ##x##-direction, there's ##F## and ##T##.
##T## acts on the pulley, which is a part of ##M_1##.
All surfaces are frictionless.
Yes, but you've missed a force. What makes M3 accelerate?
 
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Yes, but you've missed a force. What makes M3 accelerate?
The net force caused by the difference between ##N## and ##N'## (different normal reactions)?
 
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The net force caused by the difference between ##N## and ##N'## (different normal reactions)?
$$\vec{F}_{31} = -\vec{F}_{13}$$
The net force on ##M_3## is to the right, so the force ##M_3## exerts on ##M_1## must be to the left.
I guess I missed an additional constraint:
##\ddot{x}_1 = \ddot{x}_3## since ##x_3## and ##x_1## differ by a constant.
I got the right answer, thanks!
I have a question though, the equation only holds if ##M_3## is always in direct contact with ##M_1##, right?
I have not taken into account the fact that ##M_3## can swing left and right while deriving the first constraint equation.
 

haruspex

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$$\vec{F}_{31} = -\vec{F}_{13}$$
The net force on ##M_3## is to the right, so the force ##M_3## exerts on ##M_1## must be to the left.
I guess I missed an additional constraint:
##\ddot{x}_1 = \ddot{x}_3## since ##x_3## and ##x_1## differ by a constant.
I got the right answer, thanks!
I have a question though, the equation only holds if ##M_3## is always in direct contact with ##M_1##, right?
I have not taken into account the fact that ##M_3## can swing left and right while deriving the first constraint equation.
Yes, if the two do not start quite in contact then there will be a different behaviour initially, but it wouldn't last long.
 

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