Difference between revisions of "Pulley system on a double inclined plane"
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! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction | ! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction | ||
|- | |- | ||
− | | <math>m_1g</math> || [[gravitational force]] || unconditional|| <math>m_1g</math> where <math>m_1</math> is the [[mass]] and <math>g</math> is the [[acceleration due to gravity]] || vertically downward, hence an angle <math>\ | + | | <math>m_1g</math> || [[gravitational force]] || unconditional|| <math>m_1g</math> where <math>m_1</math> is the [[mass]] and <math>g</math> is the [[acceleration due to gravity]] || vertically downward, hence an angle <math>\alpha_1</math> with the normal to the incline and an angle <math>(\pi/2) - \alpha_1</math> with the incline |
|- | |- | ||
| <math>T</math> || [[tension]] || unconditional || needs to be computed based on solving equations. || pulls the mass up the inclined plane. | | <math>T</math> || [[tension]] || unconditional || needs to be computed based on solving equations. || pulls the mass up the inclined plane. | ||
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! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction | ! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction | ||
|- | |- | ||
− | | <math>m_2g</math> || [[gravitational force]] || unconditional|| <math>m_2g</math> where <math>m_2</math> is the [[mass]] and <math>g</math> is the [[acceleration due to gravity]] || vertically downward, hence an angle <math>\ | + | | <math>m_2g</math> || [[gravitational force]] || unconditional|| <math>m_2g</math> where <math>m_2</math> is the [[mass]] and <math>g</math> is the [[acceleration due to gravity]] || vertically downward, hence an angle <math>\alpha_2</math> with the normal to the incline and an angle <math>(\pi/2) - \alpha_2</math> with the incline |
|- | |- | ||
| <math>T</math> || [[tension]] || unconditional || needs to be computed based on solving equations. || pulls the mass up the inclined plane. | | <math>T</math> || [[tension]] || unconditional || needs to be computed based on solving equations. || pulls the mass up the inclined plane. |
Latest revision as of 19:56, 19 April 2012
This article discusses a scenario/arrangement whose statics/dynamics/kinematics can be understood using the ideas of classical mechanics.
View other mechanics scenarios
This article is about the following scenario. A fixed triangular wedge has two inclines and
making angles
and
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
and
, resting on the two inclines
and
respectively. The string is inextensible. The coefficients of static and kinetic friction between
and
are
and
respectively. The coefficients of static and kinetic friction between
and
are
and
respectively. Assume that
and
.
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.
Contents
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 |
---|---|---|
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The system remains at rest | ![]() |
Basic components of force diagram
There are two force diagrams of interest here, namely the force diagrams of the masses and
.
There are five candidate forces on each mass. We describe the situation for below:
Force (letter) | Nature of force | Condition for existence | Magnitude | Direction |
---|---|---|---|---|
![]() |
gravitational force | unconditional | ![]() ![]() ![]() |
vertically downward, hence an angle ![]() ![]() |
![]() |
tension | unconditional | needs to be computed based on solving equations. | pulls the mass up the inclined plane. |
![]() |
normal force | unconditional | adjusts so that there is no net acceleration perpendicular to the plane of the incline | outward normal to the incline |
![]() |
static friction | no sliding | adjusts so that there is no net acceleration along the incline. At most equal to ![]() |
The direction could be either up or down the incline, depending on whether the remaining forces create a net tendency to pull ![]() |
![]() |
kinetic friction | sliding | ![]() ![]() |
opposite direction of motion -- hence down the incline if sliding up, up the incline if sliding down. |
A similar description is valid for :
Force (letter) | Nature of force | Condition for existence | Magnitude | Direction |
---|---|---|---|---|
![]() |
gravitational force | unconditional | ![]() ![]() ![]() |
vertically downward, hence an angle ![]() ![]() |
![]() |
tension | unconditional | needs to be computed based on solving equations. | pulls the mass up the inclined plane. |
![]() |
normal force | unconditional | adjusts so that there is no net acceleration perpendicular to the plane of the incline | outward normal to the incline |
![]() |
static friction | no sliding | adjusts so that there is no net acceleration along the incline. At most equal to ![]() |
The direction could be either up or down the incline, depending on whether the remaining forces create a net tendency to pull ![]() |
![]() |
kinetic friction | sliding | ![]() ![]() |
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 , taking the component perpendicular to the inclined plane
, we get:
For , taking the component perpendicular to the inclined plane
, we get:
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
down and
up, i.e., this is what would happen if there were no static friction. Thus, the static friction
acts up the incline
and the static friction
acts down the incline
.
- The system has a tendency to slide
up and
down, i.e., this is what would happen if there were no static friction. Thus, the static friction
acts down the incline
and the static friction
acts up the incline
.
Let's consider the first case, i.e., down and
up. We get the equation for
:
and
We get a similar equation for :
and
Add (2.1) and (2.3) and rearrange to get:
Plugging in (2.2) and (2.4) into (2.5), we get:
Plugging in (1.1) and (1.2) into this yields:
Cancel from all sides to get:
This is the necessary and sufficient condition for the system to have a tendency to slide down and
up, but to not in fact slide.
Simialrly, in the other not sliding case (i.e., the system has a tendency to slide down,
up), we get the following necessary and sufficient condition:
Overall, for the no sliding case, we get the necessary and sufficient condition:
Note the following: in all these cases, it is not possible, using these equations, to determine the values of ,
and
individually.
Components along the respective inclined planes assuming sliding
We consider two cases:
-
is sliding (and accelerating) down and
is sliding (and accelerating) up, so the force of kinetic friction
acts up along
and the force of kinetic friction acts down along
.
-
is sliding (and accelerating) up and
is sliding (and accelerating) down, so the force of kinetic friction
acts down along
and the force of kinetic friction acts up along
.
We first consider the down,
up case. Let
denote the magnitude of acceleration for
.
is also equal to the magnitude of acceleration for
. We get:
with
and
with
Add (3.1) and (3.3), plug in (3.2),(1.1) and (3.4),(1.2) to get:
Rearranging, we get:
Since by our sign convention, this case is valid if:
and in particular:
The other case ( down,
up) occurs if the inequality sign is reversed, and we get, in that case, that: