1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work on Incline Planes

  1. Nov 2, 2007 #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-driven cable. (a) How much work is required to pull him a distance of 60.0m up a 30.0 degree slope (assumed frictionless) at a constant speed of 2.00m/s?

    [tex] Given:[/tex]

    [tex] \Delta r = 60m [/tex]

    [tex] \theta = 30^\circ [/tex]

    [tex] Mass = 70kg [/tex]


    2. Relevant equations
    [tex] W=F\Delta r\cos(\theta) [/tex]


    3. The attempt at a solution
    I've solved the problem, but not the way I'd like to. You may notice that, with all given data, that the triangle drawn from this would be a 30/60/90, which lets you solve it without using the work formula. You could also do it with potential / kinetic energy. This is how my horrible (would require its own thread; it's cal based physics, he doesn't teach any of the calculus aspect) physics professor did it, not bothering to showing us how to solve it with the work formula, which is what the point of the problem was, as that's what the chapter is on.

    When he gave us the work formula, he wrote it on the board and gave us one problem dealing with a level plane, said the [tex]\cos(\theta)[/tex] would be 1 because the angle was [tex]90^\circ[/tex] (he didn't explain this), and said to ignore it; that's it. Nobody in my class knows how to deal with incline planes using the work formula.

    This angers the hell out of me because I want to major in physics, and I'm going to have to end up setting myself back a semester by either passing this class and auditing/retaking a better professor's class on a branch campus, or taking this grade and teaching myself.

    After reading online, I found a spot where someone got a force by doing [tex]\sin(30^\circ)(70kg)[/tex], cutting the weight to [tex]35kg[/tex] (they said because the slope supports some of the weight). They proceeded to use the work formula, but without the [tex]\cos(\theta)[/tex] part. The final answer ends up being [tex]20580J[/tex], but I don't know WHY.

    Long story short, can someone PLEASE explain this to me? I would be forever grateful.
     
  2. jcsd
  3. Nov 2, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I'll take a shot at it. In the defining equation for work (for constant force, at least), [itex]W = F \Delta r \cos\theta[/itex], the angle theta is the angle between the force and the displacement. In this problem, you are pulling parallel to the incline and the speed is constant, so you know that the applied force must be just enough to balance the component of the weight parallel to the incline. Thus the applied force is [itex]F = mg\sin\theta[/itex] acting up the incline. Since the displacement is also up the incline, the angle between force and displacement is zero, and cos(0) = 1. Thus [itex]W = F \Delta r \cos\theta = mg\sin\theta \Delta r[/itex].

    Make sense?
     
  4. Nov 2, 2007 #3
    Thanks for the speedy reply!

    I'm mostly quiet in class, so I don't call him on his many mistakes, but it threw me for a loop when he explained "cos(90)=1 so ignore it." I imagine everyone else in the class felt the same way as I did, so kept quiet. (he gets irritable and takes it out on the class)

    I think I understand what you're saying about the angle in the [tex]\cos(\theta)[/tex] referring to the angle between force and displacement, but I'm not exactly sure why you use [tex]sin(\theta)[/tex] on the mass of the object.

    Would the triangle made of vectors be something like

    [tex]\nearrow\uparrow[/tex]
    [tex]\rightarrow[/tex]

    or would it even have anything to do with vectors?

    edit: i felt i should put a note here that after i used doc al's link, i can safely say it doesn't have anything to do with a vector triangle, just a diagram or the forces on the object.
     
    Last edited: Nov 2, 2007
  5. Nov 2, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    You need to understand how to find the components of the weight (a vector) both parallel and perpendicular to the incline surface. This might help: Forces in Two Dimensions
     
  6. Nov 2, 2007 #5
    Wow, he never taught us any of this... I want my money back.

    edit: i should say, he never taught us how to determine forces on an incline plane, only flat.
     
  7. Nov 2, 2007 #6
    Okay, I worked through all of the problems on that page, and now I understand exactly the reasons behind the [tex]\sin(\theta)[/tex]
    Thanks so much, any chance you can teach my class in his stead?
     
    Last edited: Nov 2, 2007
  8. Nov 2, 2007 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Excellent!
     
  9. Nov 2, 2007 #8
    Actually, I just realized something. All of the problems on that page deal with an object moving down the incline plane, and so do all of the diagrams. Since this problem involves the parallel force moving up the plane, drawing the [tex]F_\|[/tex] and the [tex]F_\bot[/tex] in relation to [tex]F_g[/tex] doesn't work... Since it's frictionless and at a constant speed, would [tex]F_f = F_\|[/tex]? Basically, I'm wondering how to draw the diagram :(

    For anyone else that's having problems, I did some more looking and found this:

     
    Last edited: Nov 2, 2007
  10. Nov 3, 2007 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Sure it does! In order to properly draw the free body diagram, you must know how to find the components of the weight parallel and perpendicular to the incline.

    If by [tex]F_f[/tex] you mean the applied force up the incline, then absolutely.

    Reread what I said earlier:
     
  11. Nov 3, 2007 #10
    Okay, so on an incline plane, [tex]F_\|[/tex] is always down the slope, parallel to it, because it's a component of Fg, and there is an applied force, [tex]F_a[/tex], moving up the slope? And since it's a constant speed, [tex]F_\|=F_a[/tex]. Okay. I think my physics professor has done more harm than good to my physics education, as he only taught us to determine forces on a level surface, and he always said that frictional force, [tex]F_f[/tex], was the force opposing [tex]F_\|[/tex], or as he called it, parallel force, [tex]F_p[/tex]. Therefore, I thought it was somehow the same on the incline plane, but I see now that "parallel" and "perpendicular" are components of [tex]F_g[/tex], and always go the same way, and to counteract it, there must be an applied force moving opposite the parallel. I don't think the words "applied force" ever came out of his mouth, though.

    What would I do if it weren't frictionless, though? Add the frictional force and the parallel force together, since they would both be moving opposite the applied?

    Also, if I do [tex]F=mg\sin(\theta)+(ma)[/tex] for acceleration, where would I get an [tex](ma)[/tex] equal to anything other than 0 if [tex]F_\|=F_a[/tex]. That would make Fnet be 0, then [tex]F=ma[/tex] would rearrange to [tex]a=\frac{F}{m}=\frac{0}{70kg}=0[/tex]
     
    Last edited: Nov 3, 2007
  12. Nov 3, 2007 #11
    Yes, that is exactly what you would do! And if there was an acceleration up the slope(and parallel to it) as well as friction opposing the motion,then the applied force would equal the sum of the friction force, the component of gravity parallel to the ramp and the net force. The net force would be found via m times a. For your original problem, it is worth noting that the work calculated, which you noted can also be found from the change in grav. PE, is the minimum work required. Any resisting forces such as friction and air resistance will result in a larger amount of work done and energy expended. Increasing the speed will add to the work as well. Perhaps your instructor has avoided using the term "applied force" because it sometimes causes some confusion with novices. The applied force in the case of the original problem would be the tension in the cable. For a car going up a hill, or drivng on a flat road, the applied force is a friction force in the same direction as the motion. So the applied force is the force that is the main cause of the motion. It is applied, perhaps with the help of a machine, due to the intent of a human to make an object start to move or continue moving. An object in free fall only has one force on it - gravity. But in this case, the weight or force of gravity is not typically referred to as "the applied force". But if I kick a soccer ball, the force I exert on it with my foot is an "applied force". If Big Papi hits his bat against a ball, hopefully resulting in a homerun, the force of the bat on the ball is an applied force. The force that the air exerts upward on the rotor of a helicopter (reaction force to the rotor pushing down on the air), might also be called an applied force.
     
  13. Nov 3, 2007 #12
    I guess I'm not fully understanding how you'd find acceleration if you do:
    Would the acceleration have to be given?
     
  14. Nov 3, 2007 #13

    Doc Al

    User Avatar

    Staff: Mentor

    For an object on an incline, its weight can always be broken into components parallel and perpendicular to the surface. To find the object's acceleration, use Newton's 2nd law. First you need to identify all the forces. In the special case that you know the acceleration is zero, Newton II tells you that the net force must be zero:
    [tex]\Sigma F_\| = W_\| + F_a = -mg\sin\theta + F_a = 0[/tex]

    If there's friction as well as an applied force, then you have three forces to add up. And that sum must equal ma:
    [tex]\Sigma F_\| = W_\| + F_a - F_f = -mg\sin\theta + F_a - \mu_k mg\cos\theta= ma[/tex]

    Note that I assume that the object is being dragged up the incline, thus friction points down the incline and is equal to [itex]\mu_k N = \mu_k mg\cos\theta[/itex]. N is the normal force, which must equal the perpendicular component of the object's weight. Note that I also assume that the applied force is parallel to the incline. Of course it doesn't have to be. (If it's anything other than parallel, the normal force will be affected.)

    I recommend that you work your way through some of these problems: Standard Newton's Laws Problems
     
  15. Nov 3, 2007 #14
    I understand that. So far, to find [tex]F_a[/tex], I've had to assume that [tex]F_a=W_\|[/tex]. So if they're not equal, how would I find [tex]F_a[/tex]? Basically, I only know how to do these when they're at a constant speed, sliding down the incline, or there's a pulley system with another mass and tension between them. Would I have to use a motion formula with an already given velocity and position?
     
    Last edited: Nov 3, 2007
  16. Nov 4, 2007 #15
    If someone could answer that last question about [tex]F_a[/tex] and acceleration, I'd be grateful. It's the last piece of the puzzle for me.
     
  17. Nov 4, 2007 #16

    Doc Al

    User Avatar

    Staff: Mentor

    That's true when the given acceleration is 0 and the applied force is parallel to the incline.

    It depends on the problem. Either you are given all the forces and thus you can determine the acceleration, or you're given the acceleration and must determine the forces.

    In every case you must (1) identify and analyze the forces involved, and (2) apply Newton's 2nd law.
     
  18. Nov 4, 2007 #17
    So lets say instead of moving at a constant speed in the original problem, it was accelerating, and asked me to find the acceleration. I wouldn't have enough information to do that, right?
     
  19. Nov 4, 2007 #18

    Doc Al

    User Avatar

    Staff: Mentor

    Rather than talk in generalities, dig up a specific problem that we can analyze.
     
  20. Nov 4, 2007 #19
    I have to go do a few things, but I will definitely try to find and post a problem where you have to find the acceleration up an incline of a mass. I'll try to get it as close to the original as I can. Thanks again for dealing with me and my endless questions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Work on Incline Planes
  1. Inclined plane work (Replies: 4)

Loading...