# Difference between revisions of "Pulley system on a double inclined plane"

This article discusses a scenario/arrangement whose statics/dynamics/kinematics can be understood using the ideas of classical mechanics.
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This article is about the following scenario. A fixed triangular wedge has two inclines $I_1$ and $I_2$ making angles $\alpha_1$ and $\alpha_2$ with the horizontal, thus making it a double inclined plane. A pulley is affixed to the top vertex of the triangle. A string through the pulley has attached at its two ends blocks of masses $m_1$ and $m_2$, resting on the two inclines $I_1$ and $I_2$ respectively. The string is inextensible. The coefficients of static and kinetic friction between $m_1$ and $I_1$ are $\mu_{s1}$ and $\mu_{k1}$ respectively. The coefficients of static and kinetic friction between $m_2$ and $I_2$ are $\mu_{s2}$ and $\mu_{k2}$ respectively. Assume that $\mu_{k1} \le \mu_{s1}$ and $\mu_{k2} \le \mu_{s2}$.

We assume the pulley to be massless so that its moment of inertia can be ignored for the information below.

## Summary of cases starting from rest

Case What happens qualitatively Magnitude of accelerations
$\! m_1\sin \alpha_1 - m_2\sin \alpha_2 > \mu_{s1}m_1\cos \alpha_1 + \mu_{s2}m_2\cos \alpha_2$ $m_1$ slides downward and $m_2$ slides upward, with the same magnitude of acceleration $\! a = g(m_1 \sin \alpha_1 - m_2 \sin \alpha_2 - \mu_{k1}m_1\sin \alpha_1 - \mu_{k2}m_2 \sin\alpha_2)/(m_1 + m_2)$.
$\! m_2\sin \alpha_2 - m_1\sin \alpha_1 > \mu_{s1}m_1\cos \alpha_1 + \mu_{s2}m_2\cos \alpha_2$ $m_2$ slides downward and $m_1$ slides upward, with the same magnitude of acceleration $\! a = g(m_2 \sin \alpha_2 - m_1 \sin \alpha_1 - \mu_{k1}m_1\sin \alpha_1 - \mu_{k2}m_2 \sin\alpha_2)/(m_1 + m_2)$.
$\! |m_1 \sin \alpha_1 - m_2 \sin \alpha_2| \le |\mu_{s1}m_1 \cos \alpha_1 + \mu_{s2}m_2 \cos \alpha_2|$ The system remains at rest $0$