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Water Skier Problem

  1. Nov 3, 2004 #1
    A 92 kg water skier is being pulled at a constant velocity. The horizontal pulling force is 380 N.

    The question is asking for what is the total resistive force exerted on the skier by the water and air.

    I am stumped at the moment how to calculate that if anyone can help.
     
  2. jcsd
  3. Nov 3, 2004 #2

    Doc Al

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    What does that statement allow you to conclude about the net force on the skier?
     
  4. Nov 3, 2004 #3

    Tide

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    The keywords are "constant velocity" meaning there is no acceleration! What does that tell you about the forces acting on the skier?
     
  5. Nov 3, 2004 #4
    That means acceleration = zero right? But that doesnt help me because in F=ma, if a = 0 then (0)(92) = 0 for Force, and thats not correct.
     
  6. Nov 3, 2004 #5
    I need to figure out the amoutn of friction thats what the problem is asking for i think
     
  7. Nov 3, 2004 #6

    Doc Al

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    Actually it is correct, but you have to know what it means. The F in F = ma stands for the net force on the object. Just because the net force is zero, doesn't mean no forces are acting---it just means they add to zero.

    Identify the (horizontal) forces that act on this skier.
     
  8. Nov 3, 2004 #7
    Well it says theres 380 N horizontal pulling force, is that 380 net force? Meaning including Frictional Force and The applied force?
     
  9. Nov 3, 2004 #8

    Doc Al

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    That 380 N is not the net (meaning total) force, it's only one force. (We already know the net force is zero, right?)

    Here's a big hint: If there's a force of 380 N pulling one way, what must be the magnitude and direction of the resistive force (the only other force) if they must balance out---add to zero?

    When it finally clicks you'll think that it couldn't be that easy. But it is. :smile:
     
  10. Nov 3, 2004 #9
    Or would that mean if the skier is moving at constant velocity that they two forces are at equilibrium and the frictional force would be 380 also or -380?
     
  11. Nov 3, 2004 #10

    Doc Al

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    Absolutely!
     
  12. Nov 3, 2004 #11
    Ahh that is very easy I got the answer now. Thanks a lot
     
  13. Nov 3, 2004 #12
    Well I'm having trouble with a harder one than that if your up for helping :). I know I probably have a minor problem with it. Heres the question.

    A 292-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 34.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.880, and the log has an acceleration of 0.700 m/s2. Find the tension in the rope.

    Ok heres what I got. I got myself a diagram drawn and I have the mg force exerted on the log going down which is (292*9.8) = 2714.6 which gives me my right side of my triangle. And the angle of incline is at 34 and I put the that angle in the top portion of my triangle because the angles are similar. So i've done Tan(22) * 2714.6 to find the bottom portion of the traingle or F(parallel) which I got as 1096.77. So that is the downward force acting on the Log. I also need to find the Frictional Force, which i got as 2518.21 (which is from the equation Fk=MkFn). And so those together are the forces acted down on the log. So my final equation i came up with is
    ( Ft-(Fk-Fparallel) ) / 292 = 0.700 .... ( Ft - 4448.98 ) / 292 = 0.700 and I got Ft to be 4653.38. However, thats not right. Is there something I did wrong, and I hope you can understand my work its hard typing it out here.
     
  14. Nov 3, 2004 #13

    Doc Al

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    One problem I see is that you messed up in finding the parallel and perpendicular components of the log's weight. Rethink those triangles: The weight (mg) is the hypotenuse of the triangle, the components are the other two sides. The component of the weight parallel to the incline is [itex]mg sin\theta[/itex], perpendicular to the incline (equal to the normal force) is [itex]mg cos\theta[/itex].
     
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