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Forces on a hippo Question

  1. Dec 19, 2005 #1
    I feel like i basically know how to do this question but im not 100% sure

    A 1250 kg sliperry hippo slides down a mud covered hill included at an angle at 18 degrees to the horizontal. a) if the coefficient of sliding friction between the hippo and the mud is .09, what force of friction impedes the hippo's motion down the hill? b) if the hill were steeper, how would this affect the coefficient of sliding friction?

    for a i think im suppose to do (sin 18)(1250kg x 9.8m/s^2)(.09)= 347.64N but im not sure if im suppose to use cosine

    for b i think htecoefficient of sliding friction would remain the same but im not 100%

    can someone help?
  2. jcsd
  3. Dec 19, 2005 #2


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    Staff: Mentor

    Friction is proportional (by virtue of the coefficient of friction) to the 'normal' force applied to the surface.

    The weight (mg) of an object points downward (and perpendicular to the horizontal).

    One needs to determine the normal force of the hippo on the slope.

    What happens to the normal force when the angle increases?
  4. Dec 19, 2005 #3
    am i right?

    i may be wrong but as the angle increases does the normal force decrease there for teh proper equation to use is
    (cos 18)(1250kg x 9.8m/s^2)(.09)= 1048.54N
  5. Dec 19, 2005 #4
    That's correct!
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