Question

Find accl of $m_1 \& m_3$?

Answer 1

The acceleration of $m_1$ is $9 \, m/s^2$ down the $53^\circ$ incline, and the acceleration of $m_3$ is $11 \, m/s^2$ down the $37^\circ$ incline.

Answer 2

The problem describes a pulley system with three masses: $m_1$ on a $53^\circ$ incline, $m_3$ on a $37^\circ$ incline, and $m_2$ accelerating upwards at $10 \, m/s^2$.

A common constraint relation for such pulley systems is $a_1 + a_3 = 2 a_2$, where $a_1$ and $a_3$ are accelerations down their respective inclines, and $a_2$ is the upward acceleration of $m_2$.

Given $a_2 = 10 \, m/s^2$, we have $a_1 + a_3 = 2 \times 10 \, m/s^2 = 20 \, m/s^2$.

Assuming $m_1 = m_3 = m$ (a common implication in such problems when masses are not specified) and $T$ is the tension in the string, we apply Newton's second law along the inclined planes:

For $m_1$: $T - m g \sin(53^\circ) = m a_1$ For $m_3$: $T - m g \sin(37^\circ) = m a_3$

Subtracting the second equation from the first: $m g (\sin(37^\circ) - \sin(53^\circ)) = m (a_1 - a_3)$ $g (\sin(37^\circ) - \sin(53^\circ)) = a_1 - a_3$

Using approximations: $g \approx 10 \, m/s^2$, $\sin(53^\circ) \approx 0.8$, $\sin(37^\circ) \approx 0.6$. $10 (0.6 - 0.8) = a_1 - a_3$ $-2 = a_1 - a_3$

We now have a system of linear equations:

1. $a_1 + a_3 = 20$
2. $a_1 - a_3 = -2$

Adding (1) and (2): $2a_1 = 18 \implies a_1 = 9 \, m/s^2$ (downwards along $53^\circ$ incline)

Substituting $a_1$ into (1): $9 + a_3 = 20 \implies a_3 = 11 \, m/s^2$ (downwards along $37^\circ$ incline)

Therefore, the acceleration of $m_1$ is $9 \, m/s^2$ and the acceleration of $m_3$ is $11 \, m/s^2$.