Force applied to incline plane

In summary: The force diagram on the right is similar to the one you provided, but the gravitational force is replaced with a force due to the incline. The gravitational force is still mg, but it is now divided between the two forces.
  • #1
Addem
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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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  • #3
Trying to make it appear, but it doesn't seem to be working. I've added a link, maybe that will work.
 
  • #4
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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  • #5
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.)
 
  • #6
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.
 
  • #7
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?
 
  • #8
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.
 
  • #9
Alright, I think I solved this:

https://drive.google.com/file/d/0B-HFS9bOMNAcZEtpTG5sUWlSSXM/edit?usp=sharing
 
  • #10
Compare your diagram to this one:

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

1. How does the angle of an incline plane affect the force needed to move an object?

The greater the angle of the incline plane, the more force is required to move an object up the incline. This is because the component of the object's weight acting parallel to the incline increases as the angle increases.

2. How is the force applied to an incline plane calculated?

The force applied to an incline plane can be calculated using the formula: Force = mass x acceleration. The acceleration in this case is determined by the angle of the incline plane and the force of gravity.

3. What is the relationship between the weight of an object and the force needed to move it up an incline plane?

The weight of an object is directly proportional to the force needed to move it up an incline plane. This means that the greater the weight of the object, the more force is required to move it up the incline, regardless of the angle.

4. How does friction affect the force applied to an object on an incline plane?

Friction acts in the opposite direction of the applied force on an object on an incline plane. This means that friction reduces the amount of force needed to move an object up the incline, as it counteracts the force of gravity pulling the object down the incline. However, friction can also make it more difficult to move an object up the incline by increasing the force needed to overcome it.

5. Can the force applied to an incline plane be greater than the weight of the object?

Yes, the force applied to an incline plane can be greater than the weight of the object. This is because the applied force is not only determined by the weight of the object, but also by the angle of the incline and the force of friction. If the angle is steep enough or the object is on a surface with very little friction, the force applied can exceed the object's weight.

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