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A block lies on an inclined plane with an angle of elevation

  1. Apr 10, 2017 #1
    1. The problem statement, all variables and given/known data
    A block lies on an inclined plane with an angle of elevation theta. The inclined plane is frictionless, and the plane is accelerated to the left such that the block's height remains constant. What is the net force on the block?

    I'm having trouble figuring out how to draw the correct force diagram for this. When you decompose both gravity and the force from the moving plane, is there ever a force up the ramp that balances out of the force down the ramp due to gravity?

    mass, acceleration of inclined plane, theta
    the answer is in terms of these variables

    2. Relevant equations
    F = ma
    Fblock = mg
    Fparallel = mgsintheta
    Fperpendicular = mgcostheta

    3. The attempt at a solution
    I know that the force of gravity needs to be balanced out but an upward force. But I don't know where is should be coming from. The only two forces here are gravity and force from the inclined plane.

    Thanks for the help,
    mistalin
     
    Last edited by a moderator: Apr 10, 2017
  2. jcsd
  3. Apr 10, 2017 #2

    kuruman

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    You can answer this by just looking at it. If the block were glued to the incline, the motion would look the same, would it not? What is the net force on the glued block?
     
  4. Apr 10, 2017 #3
    I don't see how this situation is the same because the ramp is frictionless and the only reason the block stays on is because the moving plane moves the block upwards. But whenever I try drawing a force diagram, there is no force vector pointing upwards.
     
  5. Apr 10, 2017 #4

    kuruman

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    Suppose you had two identical inclines with identical blocks on them, both accelerating horizontally with the same acceleration. You are told that one of them is on a frictionless incline and the other is glued to the incline. Would you be able to tell which is which? If not, then the net force on each block is the same. Remember Newton's 2nd law.

    Setting that aside, if you show me your force diagram, I will be able to troubleshoot it. You still need to draw these correctly.
     
  6. Apr 10, 2017 #5
    IMG_20170410_140325.jpg

    I tried drawing it here, but it has to be wrong because the block should have a force upwards to counteract gravity.
     
  7. Apr 10, 2017 #6

    kuruman

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    Your Fplane is way too long. There are two forces acting on the block (1) Gravity straight down and (2) Incline force, perpendicular to the incline because there is no friction. The incline force has two components (a) a vertical component that balances the weight and (b) a horizontal component that provides the acceleration a.

    See how it works?

    On edit: After seeing the original question, I see that the acceleration is not a given quantity. Your original post implied that it was, that is why I went through the "glued" block argument. Obviously, if the acceleration is given, then the net force is just mass times acceleration.
     
  8. Apr 10, 2017 #7
    Oh, so is my mistake the direction of the inclined plane force? While I drew it to be completely horizontal, the force of the inclined plane on the block is actually perpendicular to the inclined plane's surface?
    It's kind of hard for me to understand why that is the case. Since the acceleration is to the left, wouldn't ever point on the plane have a force pushing to the left on the block as well? And where is the part that counteracts the parallel component of gravity?
     
  9. Apr 10, 2017 #8

    kuruman

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    Please reread post #6. Then redraw your diagram with only two forces, gravity mg and incline FN. Draw conventional xy axes along the horizontal and vertical. Note that gravity has only a negative y-component but FN has both x and y components. Apply Newton's second law in both x and y directions.

    Perhaps you have missed the point that the incline force is a contact force which means that it adjusts itself to provide the observed acceleration. Here it must be perpendicular to the incline because we are told there is no friction. Only its length can be adjusted not, its direction. You need to adjust its length so that the weight is balanced by its vertical component. Once that is done, the horizontal component is the net force on the block responsible for the horizontal acceleration.
     
  10. Apr 10, 2017 #9
    Yeah, under the assumption that it is perpendicular, the problem makes sense and I understand how to finish it. The answer key always had the correct force diagram so I know what it looks like, I just drew mine to emphasize my thought process.

    Can you elaborate on: "Here it must be perpendicular to the incline because we are told there is no friction."
    Your explanation of how FN adjusts itself to create the acceleration makes sense, but I still don't see how it being frictionless implies a perpendicular force.

    Also, is the magnitude of FN a sum of the component of gravity and Fincline?

    I get how the vertical component of FN, cancels out with Fg, but I'm still uneasy about the fact that there doesn't seem to be a force parallel to the ramp, pointing upwards. Can that force be decomposed from FN?

    Thanks
     
  11. Apr 10, 2017 #10

    kuruman

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    A surface generates a single force on a mass placed on it. That single force can be resolved into two components, one perpendicular and one parallel to the surface. The component perpendicular to the surface is also called the normal force . The component parallel to the surface is also called the friction force . They are components, not separate entities. When a surface is "frictionless" that's another way of saying that it is unable to exert a force parallel to its surface so all that's left is the perpendicular component.

    As I just said, FN is a component of Fincline.
    It's a bad idea to create non-existent forces. There are two objects interacting with the block, Earth and Incline. Once you draw and correctly take these into account, that's all there is.
     
  12. Apr 10, 2017 #11
    Ah, thank you! That's really clear explanation.

    "The component parallel to the surface is also called the friction force."
    -I thought the friction force would be in response, but in the opposite direction, to the parallel component of the plane. So while a Ffriction does not exist, the component of the plane pointing down is still there.
     
  13. Apr 10, 2017 #12

    kuruman

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    Suppose you place a small book on top of a large book on a table and you push with your hand only on the large book. Both books will accelerate forward. What force accelerates the small book? Answer: The force of static friction coming from the large book. It is in the forward direction because that is what is needed to provide the observed (forward) acceleration. Static friction does not necessarily oppose the motion, but kinetic friction does. It's a subtle point that is often glossed over by the crass generalization that "Friction always opposes motion."
     
  14. Apr 10, 2017 #13
    True, but what makes Ffriction the same as Fparallel of the plane? Both forces do act in the same direction, but that doesn't mean that they are the same force. Yes, Ffriction is 0 because the coefficient of friction is 0, but that doesn't imply the force of the parallel component is zero. If a force pushed horizontally on that small book, even though there is no friction, the small book would still be accelerated right?
     
  15. Apr 10, 2017 #14

    kuruman

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    It seems you are still confused. Consider the small book lying on top of the large one. The only objects that can exert a force on the small book are the Earth and the large book. The Earth can only exert a vertical force so it cannot accelerate the small book in the horizontal direction. Therefore, it is the horizontal component of the large book force that accelerates the small book. That horizontal component is parallel to the book's surface, therefore it can legitimately be called "static friction." Other than the large book there is nothing else that can exert a horizontal force on the small book.
     
    Last edited: Apr 10, 2017
  16. Apr 10, 2017 #15
    Ohhh, I was always under the assumption that once you decompose the force vector, its decomposed vectors must exist as a force acting on the block.

    So what you mean is: A surface moving along its own direction underneath another object, cannot possibly exist a force on that object due to the lack of friction. Thus, in the case where it is tilted, part of its movement is sliding along the bottom surface of the block, but that force doesn't move the block. Part of its movement pushes out from the surface, and that force pushes the object left and upwards. That force is FN so the net force is finding FNsintheta. and FNcostheta = mg, since it is balanced in y direction.

    So mg/costheta * sintheta = mg tan theta = net force


    Thank you for explaining
     
  17. Apr 10, 2017 #16

    kuruman

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    A decomposed force into two components can be viewed as two separate forces. That assumption is correct. That's what I meant when I wrote
    I am not sure what you mean by the following statement but your answer is correct.
     
  18. Apr 10, 2017 #17

    haruspex

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    To add to Kuruman's explanation...
    The normal force between two bodies in contact is the minimum magnitude force necessary to prevent their interpenetration. It follows that the direction of the force is perpendicular to the plane of contact. Any component parallel to that plane would imply a greater magnitude than necessary. Any parallel force generated by the contact is, by definition, the frictional force.
     
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