A body of mass m_1 = 1 kg is pulled upward by a body of mass…
A body of mass $m_1 = 1\ \mathrm{kg}$ is pulled upward by a body of mass $m_2 = 9\ \mathrm{kg}$ sliding down an incline of angle $\alpha = 60^\circ$; the rope runs over a pulley at the top. The coefficient of friction between $m_2$ and the incline is $\mu = 0.2$. Neglecting the masses of the pulley and rope, find (a) the acceleration of $m_2$, (b) the support forces (at A and B in the figure).
Explanation of Method
A body on a 60° incline pulling up a hanging body.
Step 1: (a) Acceleration
\[ m_2 a = m_2 g\sin\alpha - \mu m_2 g\cos\alpha - T, \]
\[ m_1 a = T - m_1 g. \]
\[ a = \frac{m_2 g\sin\alpha - \mu m_2 g\cos\alpha - m_1 g}{m_1 + m_2}, \]
\[ a = \frac{76.46 - 8.83 - 9.81}{10}. \]
\[a \approx 5.78\ \mathrm{m/s^2}\]
Rope tension: $T = m_1(g + a) \approx 15.6\ \mathrm{N}$.
Step 2: (b) Forces on the supports
The figure with the supports A and B is not available, so here are the forces that the incline and the pulley have to carry:
- Incline: normal force $N = m_2 g\cos\alpha \approx 44.1\ \mathrm{N}$ and friction $\mu N \approx 8.8\ \mathrm{N}$ from $m_2$.
- Pulley axle: the two rope ends pull with $T$ each, one straight down and one down along the incline, i.e. at 30° to each other: $F = 2T\cos 15^\circ \approx 30.1\ \mathrm{N}$.
How these are shared between A and B depends on the figure.
Result
\[\boxed{\,a \approx 5.78\ \mathrm{m/s^2}\,}\]
Remark
About the original solution. Its acceleration equation is right, but it was evaluated wrongly: the numbers give 5.78 m/s², not 5.24 (and so $T = 15.6\ \mathrm{N}$, not 15.05). Its "reactions" (44.15 N and 9.81 N) are just the normal force on $m_2$ and the weight of $m_1$; the weight of $m_1$ is not a support force here, because $m_1$ hangs on the rope and accelerates.