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Homework Help: Question about finding force

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A skier of mass 70.0kg is pulled up a slope by a motor-drawn cable. How much work is required to pull the skier 60.0m up a 35 degree slope (assume to be frictionless) at a constant speed of 2.0m/s?
    Mass = 70.0kg
    Distance = 60.0m
    Angle = 35
    Velocity/Speed = 2.0m/s

    2. Relevant equations

    I would typically use the formula "W = (F)(D)(Cos0)....(cos0 = cos angle)
    I can't do this because I don't have the force. I have looked through all my notes, ask other people who take physics at my school and attempted to search online but I am unable to figure anything out.

    My Question is, how do i find force when given mass, distance, angle, and velocity/speed.

    3. The attempt at a solution

    Well i have looked for about 1 hour and 30 mins and I was unable to find anything. If someone could explain to me how to find force it would be greatly appreciated.

    Thank you
  2. jcsd
  3. Jan 28, 2009 #2


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    Science Advisor
    Homework Helper

    First of all, welcome to PF.

    What can you say about the acceleration?
    Draw a free body diagram and identify all the acting forces.
    These combined will give you the force.
    Last edited: Jan 28, 2009
  4. Jan 28, 2009 #3
    Actually you know the force!!

    Just answer, these things. What happens to a body which is acted upon by a net non zero force? Can it travel with uniform velocity?
  5. Jan 28, 2009 #4
    Thanks for the welcome.
    I drew a free body diagram and only acting force that I think there is would be Gravitational Force. The only way I know to find gravitational force is the way to find Gravitational Potential Energy which is mgh. In this case (70.0)(9.81)(60.0) = 41202 J

    I don't think that this is correct though. Am i supposed to take the square root of 41202?

    Or is correct..Fg = mg--Fg = (70.0)(9.81) = 690
    So would my force be 690?

    When the body is acted upon by a net non zero force you know that it is moving. Can it travel with uniform velocity, yes. But from this I am still unable to find force. I know I'm doing something wrong I'm just not quite sure yet.

    F = ma which I now know but I still cannot figure out acceleration. I could use d = vt + (1/2)at2, but I do not have time and I do not know how to find time either.

    Is there a formula to find time I could use? Cause I would be able to complete problem with that. I have 60.0 = (2.0m/s)(t) + 1/2(a)(t)2. If I put time into that equation I would be able to find acceleration correct? And if so i could plug it into F = ma to find my force and then put that force into W = (F)(60.0)(cos35) and answer my problem.

    If you could help me with finding time, or a simpler way to find acceleration I would greatly appreciate it.

    Thank You
  6. Jan 28, 2009 #5
    Hi Patrick!

    Actually you need to revise you Newton's Laws chapter again.

    The thing is that when a body is acted upon by a net non zero force, there is acceleration in the body in the direction of the force. Now what is acceleration?? Its the time rate of change in velocity? So when acceleration is a finite non zero quantity, the velocity would continuously change with time. But in our case, the velocity is maintained at 2 metre per second. So the acceleration will be zero. Therefore the net force acting on the skier is zero.

    List of the forces acting on the skier:
    1. Tension in the cable
    2. Normal from the surface of the slope
    3. Gravitational attraction (Weight)

    All these 3 forces add to zero. All you need to do is to resolve the weight along and perpendicular to the force.
    You will then be able to find the tension in the cable which is responsible for pulling up the skier. Have a look at the diagram.
    http://img209.imageshack.us/img209/1946/59309749ao6.png [Broken]
    Last edited by a moderator: May 3, 2017
  7. Jan 28, 2009 #6
    So the acceleration and force are zero. Then would I do W = (F)(D)(Cos0), W = (60.0)(.82) = 49 J, so my answer would be 49 J? I also could not see the diagram because the site that it is on (imageshack) is blocked at my school.

    Thank You
    Last edited by a moderator: May 3, 2017
  8. Jan 28, 2009 #7
    The tension=70*g*sin 35=393.5 N
    Displacement along the incline=60 m
    Work done=[tex]\vec{F}[/tex].[tex]\vec{d}[/tex]
    =393.5*60*cos 0
    =23608.4 J
    Do you need the diagram still ?
    I hope you understood this.
  9. Jan 28, 2009 #8
    Okay, the tension in the cable taking the skier up is m*g*sin0. We never were taught that formula. The displacement along the incline is the 60 m. which is from where skier say gets on chair lift and ends where he gets off chair lift. I've used the W = F * d before but it has always had the cos0 multiplied into it as well.

    When I calculated it though I got (70)(9.81)(sin35) = 390(390 because we are using sig figs and in this problem there are 2 sig figs.)

    Then (390)(60.0)(cos 0) = 23400 = 23000 J (sig figs again)

    So work = 23000 J

    One question I do have is why cos 0? where do you get 0 from? We only have used cos and sin with the angle of the problem only. Where does 0 come in?

    Thank You
    Last edited: Jan 28, 2009
  10. Jan 28, 2009 #9
    The answer was 24000 J. So Ritwik06 your calculations were correct because 23600 with sig figs goes to 24000. I still have a question about why you did cos 0 and where you got the zero from. If anyone can answer that for me(where you get the 0 for cos 0) i would appreciate it.

    Thank You
  11. Jan 28, 2009 #10
    I think, you still havent understood the FBD.
    Therefore I am uploading the pic here. But you will have to wait until the content is verified by PF. There is no other option as u cant access imageshack.us

    where [tex]\theta[/tex] is the angle between the force and displacement vectors. From the FBD you can see that The tension which is doing the work here is along the incline. As well as the body is also moving up the plane. so the angle between displacement and force vector is zero. They are parallel.


    Attached Files:

    • FBD.bmp
      File size:
      228.1 KB
  12. Jan 28, 2009 #11
    Okay that makes sense, thank you very much for you help.

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