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(101)Horizontal/Vertical forces on an object by a turntable

  1. Feb 17, 2008 #1
    [SOLVED] (101)Horizontal/Vertical forces on an object by a turntable

    I am taking PHY 101, which I am having a little trouble with. I took a physics course in high school which I really enjoyed; a lot of what I am doing is familiar, but the specifics confuse me occasionally. I know I have solved problem A correctly, but I am confused about B

    1. The problem statement, all variables and given/known data
    Joey puts a salt shaker near the edge and tries to spin the tray at a speed so that the shaker just barely goes around without slipping off. Joey finds that the shaker just barely stays on when the turntable is making one complete turn every two seconds. Joey's older sister measures the mass of the shaker to be 62 grams. She also measures the radius of the turntable to be 0.21 m, and she is able to calculate that the speed of the shaker as it successfully goes around in a circle is 0.6594 m/s.

    (a)What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?

    (b) What is the magnitude of the vertical part of the contact force on the shaker by the turntable?

    2. Relevant equations


    3. The attempt at a solution

    a=.2071 m/s^2
    F= .128N

    At first I was thinking the net force would be zero, because it is just sitting on the turntable. But, it is rotating obviously, so that isnt the case.
    I drew a force diagram (back view) of the shaker sitting on the edge of the turntable with the two forces acting on it being the force of gravity on the shaker by the earth (down) and the force on the shaker by the turntable diagonal up/left (the resultant of the frictional force[inward] (Fst) and perpendicular force[upward] (Fst)). However, I dont know how to calculate that diagonal force, which is what I am assuming to be the magnitude of net force.

    Thank you to all who help!!!
  2. jcsd
  3. Feb 17, 2008 #2
    You're asked only for the vertical component of the force in b) - so what contributes to this force? You're actually closer than you realize, i.e. you're making it harder than it is.

    Once you have the vertical component, you can just add it to the horizontal component you got in a), and the resulting vector is the diagonal force that you were wondering about, but which you weren't asked to find.
  4. Feb 17, 2008 #3
    Thank you for the quick reply!

    Ok, well the turntable must be pushing up on the shaker with the same force that the shaker is pushing down on the turntable. The mass of the shaker is .062kg. But, the question is asking for the magnitude of the NET force. Fnet of what I just listed would be 0 N, which I know isnt the answer.

    The force on the shaker and vice versa would be .062kg*9.8m/s, correct? Still though, the fact that the turntable is rotating must play into it somehow, otherwise the two forces would balance.

    From your response; the diagonal force wouldnt be the answer? It would make sense to me that the hypotenuse of the triangle created by the forces of .128N inward and .6076N downward would be the magnitude of the net force. In this case, im not taking into account the opposite .6076N (upward). Eh.. im rambling and loosing myself.

    Am I getting warm here?


    Ah hah! I was making it harder than neccesary. All it was asking for was force itself (.6076N). I still dont understand why the magnitude wouldnt be zero however.
    Last edited: Feb 17, 2008
  5. Feb 17, 2008 #4
    Are you asking why the net vertical force is zero (which you know since the shaker isn't accelerating up or down)? This is the same as for any object sitting stationary on a surface: the object exerts a downward force due to its weight, i.e. gravity, and the surface resists this with an equal and opposite upward force (which comes from the deformation of the molecular structure of the surface, if you must). The same holds true here - the net vertical force is zero, which is how you came up with the .6076 N upward force exerted by the turntable.
  6. Feb 17, 2008 #5
    No, I understand that the net force is zero. But thats exactly my question.

    The webassign question asks me what the magnitude of the vertical contact force is. I was assuming that Magnitude meant net force; but I just now realized it doesnt. The question wasnt asking for the net force, but simply the magnitude (strength) of the upward and downward forces. I get it now.
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