Pulley system on a double inclined plane

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This article discusses a scenario/arrangement whose statics/dynamics/kinematics can be understood using the ideas of classical mechanics.
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Pulleysystemondoubleinclinedplane.png

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. We also assume the string to be massless, so that the tension values at the two ends of the string are equal in magnitude.

Summary of cases starting from rest

These cases will be justified later in the article, based on the force diagram and by solving the resulting equations.

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\cos\alpha_1 - \mu_{k2}m_2 \cos\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\cos \alpha_1 - \mu_{k2}m_2 \cos\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

Basic components of force diagram

There are two force diagrams of interest here, namely the force diagrams of the masses m_1 and m_2.

There are five candidate forces on each mass. We describe the situation for m_1 below:

Force (letter) Nature of force Condition for existence Magnitude Direction
m_1g gravitational force unconditional m_1g where m_1 is the mass and g is the acceleration due to gravity vertically downward, hence an angle \alpha_1 with the normal to the incline and an angle (\pi/2) - \alpha_1 with the incline
T tension unconditional needs to be computed based on solving equations. pulls the mass up the inclined plane.
N_1 normal force unconditional adjusts so that there is no net acceleration perpendicular to the plane of the incline outward normal to the incline
f_{s1} static friction no sliding adjusts so that there is no net acceleration along the incline. At most equal to \mu_{s1}N_1. The direction could be either up or down the incline, depending on whether the remaining forces create a net tendency to pull m_1 down or pull it up.
f_{k1} kinetic friction sliding \mu_{k1}N_1 where N_1 is the normal force opposite direction of motion -- hence down the incline if sliding up, up the incline if sliding down.

A similar description is valid for m_2:

Force (letter) Nature of force Condition for existence Magnitude Direction
m_2g gravitational force unconditional m_2g where m_2 is the mass and g is the acceleration due to gravity vertically downward, hence an angle \alpha_2 with the normal to the incline and an angle (\pi/2) - \alpha_2 with the incline
T tension unconditional needs to be computed based on solving equations. pulls the mass up the inclined plane.
N_2 normal force unconditional adjusts so that there is no net acceleration perpendicular to the plane of the incline outward normal to the incline
f_{s2} static friction no sliding adjusts so that there is no net acceleration along the incline. At most equal to \mu_{s2}N_2. The direction could be either up or down the incline, depending on whether the remaining forces create a net tendency to pull m_2 down or pull it up.
f_{k2} kinetic friction sliding \mu_{k2}N_2 where N_2 is the normal force opposite direction of motion -- hence down the incline if sliding up, up the incline if sliding down.

Components perpendicular to the respective inclined planes

For more analysis of this part, see sliding motion along an inclined plane#Component perpendicular to the inclined plane

For m_1, taking the component perpendicular to the inclined plane I_1, we get:

\! N_1 = m_1g \cos \alpha_1 \qquad (1.1)

For m_2, taking the component perpendicular to the inclined plane I_2, we get:

\! N_2 = m_2g \cos \alpha_2 \qquad (1.2)

Note that this part of the analysis is common to both the sliding and the no sliding cases.

Components along the respective inclined planes assuming no sliding

Note that the no sliding case has two subcases:

  • The system has a tendency to slide m_1 down and m_2 up, i.e., this is what would happen if there were no static friction. Thus, the static friction f_{s1} acts up the incline I_1 and the static friction f_{s2} acts down the incline I_2.
  • The system has a tendency to slide m_1 up and m_2 down, i.e., this is what would happen if there were no static friction. Thus, the static friction f_{s1} acts down the incline I_1 and the static friction f_{s2} acts up the incline I_2.

Let's consider the first case, i.e., m_1 down and m_2 up. We get the equation for m_1:

m_1g\sin \alpha_1 = T + f_{s1} \qquad (2.1)

and

0 \le f_{s1} \le \mu_{s1}N_1 \qquad (2.2)

We get a similar equation for m_2:

T = m_2g\sin \alpha_2 + f_{s2} \qquad (2.3)

and

0 \le f_{s2} \le \mu_{s2}N_2 \qquad (2.4)

Add (2.1) and (2.3) and rearrange to get:

m_1g\sin \alpha_1 - m_2g\sin\alpha_2 = f_{s1} + f_{s2} \qquad (2.5)

Plugging in (2.2) and (2.4) into (2.5), we get:

0 \le m_1g\sin \alpha_1 - m_2g\sin \alpha_2 \le \mu_{s1}N_1 + \mu_{s2}N_2 \qquad (2.6)

Plugging in (1.1) and (1.2) into this yields:

0 \le m_1g\sin \alpha_1 - m_2g\sin \alpha_2 \le \mu_{s1}m_1g\cos \alpha_1 + \mu_{s2}m_2g\cos \alpha_2 \qquad (2.6)

Cancel g from all sides to get:

0 \le 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 \qquad (2.7)

This is the necessary and sufficient condition for the system to have a tendency to slide m_1 down and m_2 up, but to not in fact slide.

Simialrly, in the other not sliding case (i.e., the system has a tendency to slide m_2 down, m_1 up), we get the following necessary and sufficient condition:

\! 0 \le m_2\sin \alpha_2 - m_1\sin \alpha_1 \le \mu_{s1}m_1\cos \alpha_1 + \mu_{s2}m_2\cos \alpha_2 \qquad (2.8)

Overall, for the no sliding case, we get the necessary and sufficient condition:

\! |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

Note the following: in all these cases, it is not possible, using these equations, to determine the values of T, f_{s1} and f_{s2} individually.

Components along the respective inclined planes assuming sliding

We consider two cases:

  • m_1 is sliding (and accelerating) down and m_2 is sliding (and accelerating) up, so the force of kinetic friction f_{k1} acts up along I_1 and the force of kinetic friction acts down along I_2.
  • m_1 is sliding (and accelerating) up and m_2 is sliding (and accelerating) down, so the force of kinetic friction f_{k1} acts down along I_1 and the force of kinetic friction acts up along I_2.

We first consider the m_1 down, m_2 up case. Let a denote the magnitude of acceleration for m_1. a is also equal to the magnitude of acceleration for m_2. We get:

\! m_1g \sin \alpha_1 - f_{k1} - T = m_1a \qquad (3.1)

with

f_{k1} = \mu_{k1}N_1 \qquad (3.2)

and

T - m_2g\sin \alpha_2 - f_{k2} = m_2a \qquad (3.3)

with

f_{k2} = \mu_{k2}N_2 \qquad (3.4)

Add (3.1) and (3.3), plug in (3.2),(1.1) and (3.4),(1.2) to get:

m_1g\sin \alpha_1 - m_2g\sin \alpha_2 - \mu_{k1}m_1g\cos \alpha_1 - \mu_{k2}m_2g\cos \alpha_2 = (m_1 + m_2)a \qquad (3.5)

Rearranging, we get:

a = \frac{m_1g\sin \alpha_1 - m_2g\sin \alpha_2 - \mu_{k1}m_1g\cos \alpha_1 - \mu_{k2}m_2g\cos \alpha_2}{m_1 + m_2} \qquad (3.6)

Since a > 0 by our sign convention, this case is valid if:

m_1g\sin \alpha_1 - m_2g\sin \alpha_2 > \mu_{k1}m_1g\cos \alpha_1 + \mu_{k2}m_2g\cos \alpha_2 \qquad (3.7)

and in particular:

m_1g\sin \alpha_1 > m_2g\sin \alpha_2 \qquad (3.8)

The other case (m_2 down, m_1 up) occurs if the inequality sign is reversed, and we get, in that case, that:

a = \frac{m_2g\sin \alpha_2 - m_1g\sin \alpha_1 - \mu_{k1}m_1g\cos \alpha_1 - \mu_{k2}m_2g\cos \alpha_2}{m_1 + m_2} \qquad (3.9)