# Two masses and two pulleys

1. Mar 17, 2015

1. The problem statement, all variables and given/known data

Masses $M_1$ and $M_2$ are connected to a system of strings and pulleys as shown (I have attached an image). The strings are massless and inextensible, and the pulleys are massless and frictionless. Find the acceleration of $M_1$.

2. Relevant equations

$$\sum_{}^{} F_y = m \ddot{y}$$

3. The attempt at a solution

I think I got the right answer, but I want to know if my reasoning is correct. Now, both strings are massless, so the tension has to be the same throughout (for both strings). If the tension acting on $M_1$ is $T$, so is the tension acting on the second pulley. The second pulley does not accelerate, therefore the net force on the second pulley must be zero, and so the tension on each side (pulling the pulley down) must be $\frac{1}{2} T$. $M_2$ is in equilibrium, giving us the equation:
$$\frac{1}{2} T = M_2 g$$
$$T = 2 M_2 g$$
Now, applying Newton's second law to $M_1$:
$$T - M_1 g = M_1 \ddot{y}$$
$$2M_2 g - M_1 g = M_1 \ddot{y}$$
$$\ddot{y} = \frac{g(2M_2 - M_1)}{M_1}$$
I have two questions:
1) Why exactly is the net force on the second pulley zero? Is it because the setup makes it impossible for the pulley to accelerate? Or is it because its mass is negligible?
2) The equation for acceleration tells us that if $2M_2 > M_1$, $M_1$ will accelerate upwards. If $2M_2 < M_1$, $M_1$ will accelerate downwards. If $M_1 = 2M_2$, $M_1$ does not accelerate. If $M_1$ accelerates while the string is taut, doesn't this mean the second pulley should accelerate? And if the second pulley accelerates, doesn't this mean $M_2$ should accelerate? This contradicts my first equation, $T = 2 M_2 g$.

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2. Mar 17, 2015

### SammyS

Staff Emeritus
I see no attached image !

3. Mar 17, 2015

Oops! Sorry, I forgot to attach the image.
Anyway, I just attached the screenshot to the original post.

4. Mar 17, 2015

### SammyS

Staff Emeritus
Yes, the net force on the lower pulley is zero, because its mass is negligible.

String 1 does not necessarily have the same tension as string 2.

5. Mar 17, 2015

Update:
Okay, so I think I realized where I went wrong. This time, I considered the acceleration of the lower pulley (which is massless), trying to dodge division by zero wherever I can.
My inertial coordinate system is fixed to the ground, with the $\hat{\jmath}$ unit vector pointing upwards.
The positions of $M_1$, $M_2$, and $M_p$ (second pulley) are $y_1$, $y_2$, and $y_p$ respectively. Since the upper pulley is not accelerating, it can be considered as an Atwood machine with $\ddot{y}_1 = -\ddot{y}_p$. Now, for the second pulley, we have the constraint:
$$l = y_p + \pi R_p + (y_p - y_2)$$
Where $l$ is the length of the string (which is constant, since the string is inextensible) and $R_p$ is the radius of the pulley. Differentiating twice with respect to time we get:
$$\ddot{y}_2 = 2 \ddot{y}_p$$
Applying Newton's second law to the pulley:
$$T - M_p g - 2T' = M_p \ddot{y}_p$$
$M_p = 0$, therefore:
$$T' = \frac{1}{2} T$$
Where $T'$ is the tension in the lower string.
Applying Newton's second law to $M_2$:
$$\frac{1}{2} T - M_2 g = M_2 \ddot{y}_2 = 2 M_2 \ddot{y}_p = -2 M_2 \ddot{y}_1$$
$$T - 2M_2g = -4 M_2 \ddot{y}_1$$
Applying Newton's second law to $M_1$:
$$T - M_1 g = M_1 \ddot{y}_1$$
Combining the two equations:
$$\ddot{y}_1 = \frac{g(2M_2 - M_1)}{M_1 + 4M_2}$$
Is this correct?

6. Mar 17, 2015

### SammyS

Staff Emeritus
That looks good to me.

I read it fairly carefully, but didn't verify each and every detail.

7. Mar 17, 2015