21.1: Horizontal Block and Vertical Block

Consider the following problem: a block of mass $m_{1}$ is on a frictionless horizontal surface, and connected by a string, through a pulley, to a mass $m_{2}$ hanging vertically (Figure $\PageIndex{1}$). (We assume the string is unbreakable, unstretchable, and of negligible mass.) What is the acceleration of the system?

Figure $\PageIndex{1}$: Horizontal and vertical pulley connected by a string.

First, we recognize that the block $m_{1}$ will accelerate to the right, and block $m_{2}$ downward, with the same acceleration $a$, since the two blocks are tied together. Next, consider the forces on block $m_{1}$ : it has a weight $m_{1} g$, and is acted upon by a normal force, also of magnitude $m_{1} g$, so that the net force in the vertical direction is zero. This is as expected, since the block is not accelerating in the vertical direction. In the horizontal direction, the only force acting on $m_{1}$ is the string tension $T$. Thus for $m_{1}$, Newton's second law gives, in the horizontal direction,

$$\Sigma_{i} F_{i}=m_{1} a \quad \Rightarrow \quad T=m_{1} a$$

There are no horizontal forces acting on mass $m_{2}$, but there are two vertical forces: the upward tension $T$ (equal to the tension acting on $m_{1}$ ) and the downward weight force $m g$. Then Newton's second law for $m_{2}$, in the vertical (downward) direction, is

$$\Sigma_{i} F_{i}=m_{2} a \quad \Rightarrow \quad m_{2} g-T=m_{2} a$$

This gives us two simultaneous equations in the two unknowns $a$ and $T$. Adding the two equations will eliminate the tension $T$; we can then solve for the acceleration $a$ to find

$$a=\frac{m_{2}}{m_{1}+m_{2}} g$$

And then by Eq. $\PageIndex{1}$, the tension in the string is

$$T=\frac{m_{1} m_{2}}{m_{1}+m_{2}} g$$

Now let's consider the same problem, but this time we'll include friction acting on the horizontal block. In this case, Newton's second law for $m_{1}$ (Eq. $\PageIndex{1}$) will include a frictional force $f=\mu n=\mu m_{1} g$ (where $\mu$ is the coefficient of (kinetic) friction) acting to the left, and becomes

$$\Sigma_{i} F_{i}=m_{1} a \quad \Rightarrow \quad T-\mu m_{1} g=m_{1} a$$

Newton's second law applied to mass $m_{2}$ is the same as before:

$$\Sigma_{i} F_{i}=m_{2} a \quad \Rightarrow \quad m_{2} g-T=m_{2} a$$

Adding these two equations to eliminate the tension $T$, we find the acceleration $a$ to be

$$a=\frac{m_{2}-\mu m_{1}}{m_{1}+m_{2}} g$$

and the tension to be (using Eq. $\PageIndex{5}$),

$$T=\frac{(1+\mu) m_{1} m_{2}}{m_{1}+m_{2}} g$$

Notice that these last two equations reduce to their frictionless counterparts when $\mu=0$.