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Pulley system on a double inclined plane

, 01:29, 14 August 2011
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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 [[involves::double inclined plane]]. A [[involves::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.
{| class="sortable" border="1"
| $\! |m_1 \sin \alpha_1 - m_2 \sin \alpha_2| \le |\mu_{s1}m_1 \cos \alpha_1 + \mu_{s2}m_2 \cos \alpha_2|$ || 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:

{| class="sortable" border="1"
! 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 $\theta$ 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. Assuming the block was ''initially'' at rest, it can only slide down, hence ''up the incline'' is the only possibility.
|}

A similar description is valid for math>m_2[/itex]:

{| class="sortable" border="1"
! 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 $\theta$ 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. Assuming the block was ''initially'' at rest, it can only slide down, hence ''up the incline'' is the only possibility.
|}

===Components perpendicular to the respective inclined planes===

{{quotation|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:

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