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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?
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?
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