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Direction of friction

  1. Jan 16, 2017 #1
    1. The problem statement, all variables and given/known data
    A homogeneous semicircular disc of mass m and the radius r is released from rest in the vertical position shown in the figure. Determine a minimum value of the coefficient of friction μ between the disc and the horizontal surface that prevents sliding in the initial moment
    34f8129b52201e3af8c85c57459f5685.jpg

    2. Relevant equations
    Because it is a rigid body then:
    OG=4R/3π

    c6e5c6c3f2aa52ce1bbccbb661676137.jpg


    Force and torque equation:
    ex: m*agx = -F (where F is friction)
    ey: m*agy= N-m*g

    IG * α = N*OG - F*R

    I know how to solve rest but the only thing I really got stuck and do not understand, is the direction of the friction. What I learned early in physics was to friction used to be the opposite of the direction of movement. But I see some issues that friction has the same direction as the movement! why is it like that ?!
     
  2. jcsd
  3. Jan 16, 2017 #2

    TSny

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    You should distinguish between the direction of the motion of the center of mass of an object and the direction that a part of the surface of the object tends to slip relative to another surface. For example, if you are standing and begin to walk, the friction force on your shoe is in the same forward direction that you start moving. But as you start to walk, the bottom surface of your shoe tends to want to slip backward relative the floor (i.e., the friction opposes the tendency to slip).

    Another example would be a box sitting in the back of a truck that begins to move forward from rest. The friction force on the box is in the same direction as the motion of the box. But the bottom surface of the box tends to want to slip backward relative to the surface of the truck. So, the friction is opposing this slippage.

    Can you see that it's similar in your disk problem?
     
  4. Jan 16, 2017 #3

    mfb

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    Static friction can point in any direction. Friction is supposed to prevent moving here, so object at the point of contact does not move anyway.
     
  5. Jan 16, 2017 #4
    OK I thought that the direction of the disk is counterclockwise and the friction will be clockwise. Is this right to think so ?!
    Now it will be interesting for me when you say about box and truck....
    look at this picture :

    f0c83685608b196cd1bef7de607f9f72.jpg

    If we assume that the wheel spins at a constant speed and no other forces than normal, weight and friction affect it, so in this case we say that friction is the opposite of the direction of movement so what is the difference now between this type of example and the like you say with box in a truck?! Why the friction is in this case in opposite direction of movement?

    And if we assume the wheel as a homogeneous circular disc, I know that gravity will be right in the middle, so the friction because of it in the opposite direction?

    I will be grateful if you answer the questions that I highlighted in bold:smile:
     
    Last edited: Jan 16, 2017
  6. Jan 16, 2017 #5

    vela

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    For cases like these, I find it useful to imagine the motion of an object if friction weren't present. Consider the box in the back of the truck. With no friction, there will be no net force on the box as the truck accelerates forward, so according to the first law the box will remain at rest. As the truck accelerates forward, the box will therefore slide backward relative to the truck. If friction is present, it will oppose this sliding motion, so it will point forward.

    For the wheel on the incline, let's assume it starts from rest. If there's no friction, it will simply slide down the incline without rotating. At the point of contact with the incline, the bottom of the wheel slides down the incline, so friction on the wheel will point up the incline.

    If you want more practice, you might want to think about these situations: (1) A car stuck in the snow. The driver hits the gas pedal, causing the wheel to spin in the snow, but the car doesn't move forward. (2) A car on a slick road approaching a red light. The driver locks the brakes, so the wheels stop spinning even though the car is still moving forward. (3) A car moving normally down the road at constant velocity. In each case, which way does the force of friction acting on the wheel point?
     
  7. Jan 16, 2017 #6
    1) I thought to this situation very much but I guess the friction in this case will be in opposite direction of movement but I don't know if I'm right
    2) It is clear that friction is in the opposite direction of movement of the car
    3) As I see now and I worked with such questions, I know the direction of friction is in the same direction of movement in this case

    But I'm very grateful if you give me the right answer of situation 1
     
  8. Jan 16, 2017 #7

    vela

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    The car's not moving, so what you do mean by "the opposite direction of movement"?

    What if the wheels only partially lock so they're still rotating but not fast enough to roll without slipping? (That would have been a better question.)

    This is wrong.


     
  9. Jan 16, 2017 #8
    OK for 1 I now I see, If we assume that the wheel rotates clockwise then the friction in the point between wheel and snow will be counterclockwise if I am right ! :smile:

    for 2) I guess friction is still in the opposite direction of movement
    for 3 ) I give up! :sorry::frown:
     
  10. Jan 16, 2017 #9

    vela

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    It would be more accurate to say the torque due to the force of friction is in the counterclockwise direction.

    Can you explain your reasoning?

     
  11. Jan 16, 2017 #10
    What I understand from this question is that driving a car in a wet runway. We see the red light and trying to slow the car and when the wheel locks up, there is no rotation of the wheels. The slide is forward in this case, the friction will be in the opposite direction. But if the wheel does not lock up 100% and spins a little then I think there will be torque and rotation and friction will be in opposite direction of the movement say counterclockwise!

    But for the third situation... I am frustrated:frown:
    But I think when you say the velocity is not changing that means there is no acceleration and all forces will cancel out each other.... should I say there is no friction in this case ?!
     
  12. Jan 16, 2017 #11

    Stephen Tashi

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    If you want to drive yourself crazy consider the above picture in two other situations.

    1) The wheel is the wheel of a unicycle and the cylist is trying to accelerated down the incline plane faster than gravity would allow him to coast down.

    2) The wheel is the wheel of a unicycle and the cyclist is trying to cycle up the inclined plane.

    Would this rule work ?: The direction of the friction of a surface against a wheel opposes the net applied torque on the axle of the wheel by other forces. If there is no net applied torque on the axle of wheel due to other forces then the direction of friction points in the direction that applies a torque explaining the direction of the wheel's rotation.
     
  13. Jan 16, 2017 #12

    haruspex

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    In general, when discussing rotation, you need to be careful to identify the axis being used.
    The half disk will rotate anticlockwise about its mass centre, and likewise about its point of contact with the ground. But if you were to pick an axis more to the left it would be less clear. You would have to count the linear motion of the mass centre, and that could have a clockwise moment about the axis.
    If you mean about the point of contact, gravity has an anticlockwise moment, so can in itself explain the rotational movement. Indeed, friction has no moment about that axis.
    Using the mass centre as the axis, yes, you can discount gravity and conclude that friction acts anticlockwise about that point.
     
  14. Jan 16, 2017 #13

    vela

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    Yup! If there's no slipping and the disk moves at constant velocity, its rotational rate must also be constant. If there were a force of friction, it would result in a net torque on the disk and cause its rotation to speed up or slow down.

    Anyway, I hope with this practice you're getting a better feel to be able to do as TSny advised in his post: "distinguish between the direction of the motion of the center of mass of an object and the direction that a part of the surface of the object tends to slip relative to another surface."
     
  15. Jan 16, 2017 #14

    Stephen Tashi

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    The third situation is not what we'd call "a car moving normally down the road" from a practical point of view because in a practical situation there are many forces being applied to the wheel. Consider a theoretical situation. Suppose we spin an ideal wheel on a frictionless axel. Assume it's in a vacuum, so there is no drag from the air. The wheel will keep spinning at a constant angular velocity once the force used to spin it ceases. Now suppose we also have a theoretical treadmill that's perfectly flat and moving at a constant velocity that's exactly equal to the tangential velocity of a point on the circumference of the wheel. The wheel does not require any applied force to keep it spinning. If we move the treadmill toward the wheel so it touches the wheel "at a point", the treadmill and the wheel can each keep moving happily along at their constant velocities without any force of friction developing at the point of contact.
     
  16. Jan 17, 2017 #15
    Thank you guys ! Excuse me that I ask so much about this! but I saw two other situations that made me dizzy!
    Look at this picture (1)

    valsel.jpg

    Here we have a cylinder that will be influenced by a force P. The cylinder is at rest initially,
    The force P forms 45 degrees with the horizontal surface which is considered easily (soft) in conjunction with the cylinder. The direction of the friction between the cylinder and the surface that prevents sliding is ??!
    I see in my solution the friction has counterclockwise direction. That means this is against the direction of movement ....

    Another example which I compare with (1) and made me dizzy is this :
    (2)
    wheel.jpg

    A homogeneous wheel of radius R spinning with angular velocity ω shortly before a roadway. The wheel is released from this position and starts to move along a straight horizontal path. The wheel slips and rolls in the beginning. Determine the maximum speed v1 which is achieved by wheels center of mass.

    The only thing that I am hesitant to find is the direction of friction. I know how to solve the rest. I see in my solution that the friction in this case is in the direction of movement. Counterclockwise.
    Perhaps this question is poorly worded. From (1) and (2) I know that there is an acceleration that makes both cylinders roll forward but in (2) I see the word slip and rotation.... I know further that the wheel has the max velocity v1 when it rolls without sliding. Which means m*R*α (α is angular acceleration)....
     
    Last edited: Jan 17, 2017
  17. Jan 17, 2017 #16

    haruspex

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    Since it starts at rest, it is initially not rotating. Without friction, there would be nothing to make it start rotating. Which surfaces would slide against each other? Which way does friction then act on the roller? Will friction acting that way cause the roller to rotate counterclockwise?

    Yes, friction will exert a force to the right, producing a counterclockwise moment at the wheel's centre.
    "Slips and turns" would be better. Is that what you mean?
    Not sure what you are saying there. It means what about m*R*α?

    By the way, there is a trick for solving these problems about rolling contact velocities after an initial skid. If you consider angular momentum about a point at ground level, friction has no moment about that point, so angular momentum is conserved.
     
  18. Jan 17, 2017 #17
    OK if in (1) the cylinder does not rotate at the beginning I can imagine that the cylinder will just slides to the left and since it does not rotate the friction in this moment will be in opposite direction of movement! I wish I'm right....

    For (2) yes I see that the friction will be counterclockwise but I misunderstand the word Slips and turns, well I can imagine two things :

    1) Just a car will accelerate in a snowy way but sliding reduces its power to move forward and in this case the friction is in the same direction of movement
    2) I imagine too that the wheel just slides wheel slips as if it would be on the ice and continue to run as skating on ice. I mean drift allows it to go forward and the friction has an opposite direction but it feels I am wrong to think so
     
  19. Jan 17, 2017 #18

    haruspex

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    If the wheel does not rotate at all then it slips without turning.
    If it rotates at rate ω= v/r (in the forward direction) then it turns without slipping, i.e. rolls. The velocity of the wheel's centre is v; the velocity of the lowest point of the wheel relative to the centre is -ωr; so the velocity of the lowest point in the ground frame is v-ωr=0.
    If the wheel rotates at any other rate then it both slips and turns.
     
  20. Jan 17, 2017 #19
    proxy.php?image=https%3A%2F%2Fanonimag.es%2Fi%2F34f8129b52201e3af8c85c57459f5685.jpg

    Pardon me ! Here in first question I asked, the half cylinder was at rest too... But friction has counterclockwise direction
    and here
    valsel.jpg
    The situation is the same except that we have a whole cylinder here. Is it because the entire cylinder is affected by a force?! I see the only difference in these two questions that the first is not to be affected by a force while the other is affected by a force ... am I right now?!
     
  21. Jan 17, 2017 #20

    haruspex

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    But it is. Gravity acts on the half disc, creating a counterclockwise torque about the point of contact.

    For the cylinder on the soft surface, things are a bit complicated. There is no single point of contact to take moments about.
    Suppose instead that it is a hard surface, but in the shape shown. Now the cylinder will not move. So its movement on a soft surface must be connected with the softness somehow.
    As the force is applied, the surface to the left yields. The cylinder would tend to slide forward into the yielded space, so the friction of the surfaces in contact causes the cylinder to rotate.
     
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