How Is Force Applied to an Inclined Plane?

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SUMMARY

The discussion centers on the application of force on an inclined plane, specifically how gravitational force can be decomposed into components that act parallel and perpendicular to the incline. Newton's Laws of Motion are utilized to analyze the forces at play, with emphasis on the gravitational force (mg) and the effects of horizontal applied forces. Participants clarify the terminology, noting the importance of distinguishing between "perpendicular" and "vertical" components in force diagrams. The conclusion emphasizes that to maintain equilibrium on the incline, the forces along the incline must cancel each other out.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Knowledge of gravitational force and its components
  • Familiarity with force diagrams and vector decomposition
  • Basic principles of static equilibrium
NEXT STEPS
  • Study vector decomposition in physics
  • Learn about static equilibrium conditions on inclined planes
  • Explore examples of force diagrams in mechanics
  • Investigate the effects of friction on inclined planes
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of forces on inclined planes.

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Homework Statement



The problem is described and drawn here:

https://drive.google.com/file/d/0B-HFS9bOMNAcMjNacXlha0Z6cVk/edit?usp=sharing

Edit: The picture seems not to be showing up, so the address is: https://drive.google.com/file/d/0B-HFS9bOMNAcMjNacXlha0Z6cVk/edit?usp=sharing

Possibly this clickable link will work: https://drive.google.com/file/d/0B-HFS9bOMNAcMjNacXlha0Z6cVk/edit?usp=sharing

Homework Equations



Newton's Laws of Motion, gravitational force is mg, and the redirection of force on an inclined plane.

The Attempt at a Solution



In the picture above--basically, I just am not sure how the force of the mass is directed on the plane. I think I get the rest of the important stuff: The acceleration of the system is a, so you can use F = ma since it will only accelerate in the horizontal direction, and each individual object will also have that same acceleration, etc.
 
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Trying to make it appear, but it doesn't seem to be working. I've added a link, maybe that will work.
 
Look at your force diagram and the forces on the block: gravity is pulling it downward but the inclined plane prevents it from falling freely instead it breaks up the gravitational force into two components:

One force component should be parallel to the inclined plane and one should be vertical to the inclined plane.

The one force component parallel is the force that moves the block

The vertical force has no effect unless friction is to be considered in the problem.
 
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The force of gravity, which is straight down, can be converted to a component parallel to the inclined plane and one that is perpendicular to it. It is the component perpendicular to the plane that acts on it. Further, the horizontal force pushing from the side can be converted to such components and, again, it is the perpendicular component that acts on the plane.

(I agree with jedishrfu though I think his use of "vertical to the inclined plane" rather than "perpendicular to the inclined plane" can be confusing.)
 
HallsofIvy said:
The force of gravity, which is straight down, can be converted to a component parallel to the inclined plane and one that is perpendicular to it. It is the component perpendicular to the plane that acts on it. Further, the horizontal force pushing from the side can be converted to such components and, again, it is the perpendicular component that acts on the plane.

(I agree with jedishrfu though I think his use of "vertical to the inclined plane" rather than "perpendicular to the inclined plane" can be confusing.)

Yes, sorry for the poor choice of words it should have been vertical. I was trying not to give too much away.
 
HallsofIvy said:
Further, the horizontal force pushing from the side can be converted to such components and, again, it is the perpendicular component that acts on the plane.

(I agree with jedishrfu though I think his use of "vertical to the inclined plane" rather than "perpendicular to the inclined plane" can be confusing.)

I'm a bit confused about this. The applied force is purely horizontal, so as far as it acts on the plane, I would think it acts in the horizontal direction.

Do you mean that, as the plane acts on the block, the action is purely perpendicular to the incline? So the force the plane applies to the block is perpendicular to the incline, and then to relate that to the fact that the block is stationary relative to the plane, I would have to find the vertical component of that?
 
The applied horizontal force is equal to the sum of one force along the incline and one force into/out of the incline. It's easier to break up the force this way so that you can figure out the force applied to the block in the along-incline direction.

Similarly, downward gravity is equal to the sum of one force along the incline and one force into/out of the incline.

To keep the block at the same place on the incline, you need the forces along the incline to cancel each other.
 
Alright, I think I solved this:

https://drive.google.com/file/d/0B-HFS9bOMNAcZEtpTG5sUWlSSXM/edit?usp=sharing
 
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Compare your diagram to this one:

http://www.clickandlearn.org/images/incline.gif
 

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