Question

In a simple Atwood machine, two unequal masses m, and m₂ are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure 5-E7) m₁ = 300 g and m₂ = 600 g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp on the pulley. m10 m2 Figure 5-E7 23. Consider the Atwood machine of the previous problem. * The larger mass is stopped for a moment 2-0 s after the system is set into motion. Find the time elapsed before the string is tight again.

Answer 1

(a) The first block travels approximately $6.53\;m$ in 2 seconds.
(b) The tension in the string is approximately $3.92\;N$.
(c) The force exerted by the clamp on the pulley is approximately $7.84\;N$.
(d) Once $m_2$ is stopped at 2.0 s, the string becomes tight again after about $1.34\;s$.

Answer 2

We solve the problem step‐by‐step.

Step 1. Find the acceleration

For an Atwood machine the acceleration is

$$a=\frac{(m_2-m_1)g}{m_1+m_2}$$

with

$$m_1=0.3\;kg,\quad m_2=0.6\;kg,\quad g=9.8\;m/s^2.$$

Thus

$$a=\frac{(0.6-0.3)\times9.8}{0.3+0.6}=\frac{0.3\times9.8}{0.9}\approx3.27\;m/s^2.$$

Step 2. (a) Distance travelled by the first block in 2.0 s

Starting from rest with constant acceleration,

$$s=\tfrac12 a t^2=\tfrac12\times 3.27\times (2)^2\approx 6.53\;m.$$

Step 3. (b) Tension in the string

Take mass $m_1$ (which moves upward) and write Newton’s 2nd law:

$$T-m_1g=m_1a\quad\Longrightarrow\quad T=m_1(g+a).$$

So

$$T=0.3(9.8+3.27)=0.3\times13.07\approx3.92\;N.$$

(One may check using $m_2$:

$$m_2g-T=m_2a\Rightarrow T=m_2(g-a)=0.6(9.8-3.27)\approx3.92\;N.)$$

Step 4. (c) Force exerted by the clamp on the pulley

The light, smooth pulley is acted on by the two string tensions. In a standard Atwood machine the rope contacts the pulley over a semicircular arc. A standard result is that the net force on the pulley (which must be balanced by the clamp) is

$$F_{clamp}=2T\quad\Longrightarrow\quad F_{clamp}\approx2\times3.92=7.84\;N.$$

Step 5. (d) Time for the string to become tight again when $m_2$ (the larger mass) is stopped at $t=2.0$ s

At $t=2.0\;s$:

• Both masses have reached speeds $|v|=a\times2\approx6.53\;m/s$ (with $m_1$ moving upward and $m_2$ downward).
• The string is just taut.

Now suppose at $t=2.0\;s$ the larger mass $m_2$ is suddenly held fixed while $m_1$ continues upward. Immediately, $m_1$ has speed $6.53\;m/s$ upward and will continue its motion as a free projectile (ignoring the now‐slack string) under gravity alone.

1. Extra upward travel:
   The additional height $\Delta h$ gained by $m_1$ is found from

   $$0=v^2-2g\Delta h\quad\Longrightarrow\quad \Delta h=\frac{v^2}{2g}=\frac{(6.53)^2}{2\times9.8}\approx\frac{42.67}{19.6}\approx2.18\;m.$$

   This extra travel creates a slack of $2.18\;m$ in the string on the $m_1$ side.

2. Time to re‐take up the slack:
   At the top, $m_1$ has zero speed. It then falls a distance $2.18\;m$ under gravity. The time $t_{\text{fall}}$ to fall that distance is given by

   $$2.18=\tfrac12g t_{\text{fall}}^2\quad\Longrightarrow\quad t_{\text{fall}}=\sqrt{\frac{2\times2.18}{9.8}}\approx\sqrt{0.444}\approx0.67\;s.$$

3. Also, the time to reach the maximum height (from $t=2.0\;s$) is

   $$t_{\text{up}}=\frac{6.53}{9.8}\approx0.67\;s.$$

Thus the total time from $t=2.0\;s$ until the string becomes taut again is

$$t_{\text{slack}}=t_{\text{up}}+t_{\text{fall}}\approx0.67+0.67=1.34\;s.$$