Two masses and two pulleys

Masses M_1 and M_2 are connected to a system of strings and pulleys as shown. The strings are massless and inextensible, and the pulleys are massless and frictionless. Find the acceleration of M_1.

Screenshot 2015-01-25 at 11.51.26

Problem Source : An Introduction to Mechanics

We first look for the relationship between the tension of the two strings. Consider the tension force exert on the lower pulley with mass m,

T_1 - 2T_2 - mg = ma

Although there is acceleration on the pulley, it’s mass is 0. So, we can conclude that

T_1 = 2T_2

Screenshot 2015-01-25 at 12.01.55

Let the acceleration of M_1 be -a_1, then the acceleration of the lower pulley will be a_1.

Screenshot 2015-01-25 at 12

If l is the length of the string, we have

x_1 + x_2 + \pi R = l

Differentiating twice we have

\begin{aligned}  \ddot{x_1} + \ddot{x_2} &= 0 \\ (a_1 - 0) + (a_1 - a_2) &= 0 \\ a_2 &= 2a_1  \end{aligned}

Now, we have all the details needed to solve this problem, with two variables and two equations that study the forces on M_1 and M_2.

\begin{aligned}  2T - M_1g &= M_1(-a) \\ T - M_2g &= M_2(2a)  \end{aligned}

Substitute T,

\begin{aligned}  \displaystyle M_1g-M_1a &= 2M_2g+4M_2a \\ (M_1+4M_2)a &= (M_1-2M_2)g \\ a &= \bigg( \frac{M_1-2M_2}{4M_2+M_1}\bigg)g  \end{aligned}

We can also find the tension force on the string T. Substitute a on the second equation,

\begin{aligned}  \displaystyle T &= 2M_2a+M_2g \\ &= M_2\bigg( \frac{2M_1-4M_2}{4M_2+M_1} + 1 \bigg)g \\ &= \bigg ( \frac{3M_1M_2}{4M_2+M_1}\bigg )g  \end{aligned}

Two masses and two pulleys

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s