Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Force Problem~

  1. Feb 13, 2006 #1
    Holding onto a tow rope moving parallel to a frictionless ski slope, a 62.5 kg skier is pulled up the slope, which is at an angle of 8.2° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 2.00 m/s and (b) v = 2.00 m/s as v increases at a rate of 0.105 m/s2?

    when trying to figure this out, i used the formula,
    T = mg sin (theta)
    = (62.5)(9.8)(sin 8.2)
    = 87.36N

    then Fn = mg cos (theta)
    =(62.5)(9.8)(cos 8.2)
    = 606.2N

    then the magnitude came to = 612.5N

    for part B....
    I used Fn = m(g+a)
    = (62.5)(9.8 + .105)

    But both of those answers are incorrect... am i going about this problem the wrong way?? any help would be appreciated.
  2. jcsd
  3. Feb 14, 2006 #2
    For part (a): if the skier is moving at a constant velocity, then the force with which the rope pulls him up the slope is equivalent to the force with which gravity pulls him down the slope. This results in zero net force and zero net acceleration (thus, constant velocity). So, what we need to find is the component of the gravitational force that is parallel to the surface of the slope. And that was the first thing you did:
    That's completely correct. You found the component of gravity that is pulling the skier down the slope, and you set it equal to the force of the rope. You should have stopped there - 87.36 N is the answer for (a).

    Your mistake was in adding [itex]mgcos(8.2)[/itex] <-- that's the formula for normal force, that is, the component of gravity that acts PERPINDICULAR to the surface. You would have used this formula if the surface had any measurement of friction, as the force of friction = Fn * friction constant, then you would have added in that friction force. But in the case of this problem, no friction, so don't worry about normal force.

    Part (b) asks for the force of the rope if the skier is accelerating at 0.105 m/s^2. In this case, our forces obviously don't add to simply zero. However, it's not too difficult. Follow this basic idea:
    Sum of all forces = m*a
    That is, the sum of all of the forces acting upon the skier is equivalent to the mass of the skier * his acceleration (all of these along the surface of the slope).

    So, let's find the forces acting upon the skier. The obvious one is gravity acting along the slope, which is as we said in (a), [itex]mgsin(8.2)[/itex]. Our only other force acting along the slope is the force of the rope pulling the skier, which is what we want to solve for, so we'll call it F. As we've said, sum of forces = m*a:

    [itex]F - mgsin(8.2) = ma[/itex]
    [itex]F - (62.5 kg)(9.8 \frac{m}{s^2})sin(8.2) = (62.5 kg)(0.105 \frac{m}{s^2})[/itex]
    [itex]F = 93.92 N[/itex]

    Keep in mind that we needed to account for direction in this equation. F and a are pointed up the slope so we made them positive, whereas the gravitational force is pointed down the slope so we made it negative.
  4. Feb 14, 2006 #3
    thanks so much. i understand now. =)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook