Problem 230
Medium Translational Dynamics

In the system shown in diagram E, the masses are m_1 = 10 kg…

In the system shown in diagram E, the masses are $m_1 = 10\ \mathrm{kg}$ and $m_2 = 5\ \mathrm{kg}$; the coefficient of friction between $m_1$ and its surface is $\mu = 0.2$, and the incline angle is $\alpha = 30^\circ$. Find (a) the acceleration of the system and (b) the rope tension.

Explanation of Method

A body on a table pulled by a body on an incline.

Step 1: Assumed arrangement

Diagram E is not available. As in the original solution, $m_1$ lies on a rough horizontal table and is pulled over a pulley by $m_2$, which slides down a smooth incline of angle $\alpha$.

\[ m_2 a = m_2 g\sin\alpha - T, \]

\[ m_1 a = T - \mu m_1 g. \]

Step 2: (a) Acceleration

\[ a = \frac{m_2 g\sin\alpha - \mu m_1 g}{m_1 + m_2} = \frac{24.53 - 19.62}{15}. \]

\[a \approx 0.327\ \mathrm{m/s^2}\]

Step 3: (b) Tension

\[ T = m_1 a + \mu m_1 g = 3.27 + 19.62. \]

\[T \approx 22.9\ \mathrm{N}\]

$\mu m_1 g$ $m_1 = 10$ $m_2 = 5$ $30^\circ$ $a$ $a = \frac{m_2 g\sin\alpha - \mu m_1 g}{m_1 + m_2} \approx 0.327\ \mathrm{m/s^2}$
The arrangement assumed in the original solution (diagram E is not available): $m_1$ on a rough horizontal table, pulled by $m_2$ sliding down a smooth 30° incline.

Result

\[\boxed{\,a \approx 0.327\ \mathrm{m/s^2}\,}\]

\[\boxed{\,T \approx 22.9\ \mathrm{N}\,}\]

Remark

Check. For this arrangement the original results are correct. If instead $m_1$ were on the incline and $m_2$ hanging, then $m_1 g\sin\alpha = m_2 g = 49.05\ \mathrm{N}$ exactly: the pull and the slope force would balance, friction would hold everything at rest ($a = 0$, $T = 49.05\ \mathrm{N}$). The answer depends on the figure.