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  1. Oct 8, 2004 #1
    Please tell me if my answers are right or wrong. Thanks

    A box of mass 12 kg is placed on a rough plane inclined at an angle of 25 degrees from the horizontal. The coefficient of static friction between the plane and the box is 0.6, and the coefficient of kinetic friction is 0.5. To solve the following problems, choose the x-axis (upward) along the plane and y-axis normal to it.

    a) What are all the forces acting on the box.

    kinectic force, normal force, static force, and weight

    b) Calculate the normal and frictional forces.

    Normal Force = 117.72 newtons
    Kinetic Force = -58.86
    Static Force = 70.63

    c) Will the box remain stationary? Explain

    No, because the static friction (force) is higher/stronger that the kinetic friction (force).

    d) Now the man pushes it upwards with a force of 86 Newtons along the incline. What are the forces acting on the box.

    normal force, kinetic force, static force, weight, and pushing force (P)

    e) Calculate the acceleration of the box.

  2. jcsd
  3. Oct 8, 2004 #2
    Static friction and kinetic friction can't act on a particle simultaneously. if the block is in stationary, then the friction acts on it is static friction. if it is moving, then it is kinetic fraction. if the weight of the block in the y direction is greater than the max static friction, then if will move downward. if not, it will be in stationary with the static friction equals to the block weigh in the y direction.
  4. Oct 9, 2004 #3
    can someone tell me if I got b and e right?
  5. Oct 9, 2004 #4


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    First off, I'm getting a different answer for the normal force - specifically, I'm getting 106.58 N. The weight of the box is 12kg(9.8 m/s^2) = 117.6 N. Is it possible that you're calculating the weight and assuming that's the normal force? Remember - the normal force only resists motion normal (perpendicular) to the plane.

    As for the other - remember that, when you use the coefficient of static friction, you're actually calculating the maximum possible static friction. The actual value will be that or less, depending on just how much it needs to resist. Static friction will never be large enough to cause something to move - just large enough to keep it from moving.
  6. Oct 10, 2004 #5
    (a) The normal force, the frictional force (static friction specifically) and its weight.
    (b) Normal force = 107 N
    Frictional force = 49.8 N
    (c) Yes, because the maximum static friction is greater than the x component of the weight of the box and in this case, the static fiction is 49.8 N which is equal to the x component of the box’s weight and is just enough to prevent the box from moving downwards.
    (d)The normal force, the frictional force (static friction specifically), the pushing force and its weight.
    (e) 0. the pushing force of 86 N is not enough to move the box upward because the sum of the forces in the negative x direction is 49.8 N ( the x component of the weight) plus the maximum static friction ( 64.2 N) = 114 N. the pushing force must exceed 114 N for the box to move upward, then only it can be reduced a little (minimum 104 N = the x component of the weight plus the kinectic friciton) for the box to keep on moving because once it moves, we will have kinetic friction which is less than the static friction .
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