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Homework Help: Forces - friction, centripetal

  1. Nov 14, 2006 #1
    I've been doing some practice force questions and there are a few I don't understand. I have the solutions to them, so my questions will be relatively short, but I just need to have some things cleared up.

    The first one:
    "A circus performer of weight W is walking along a 'high wire' as shown in the diagram below (first attachment). The tension of the wire is:
    a.) Approximately W
    b.) Approximately W/2
    c.) Much less than W
    d.) Much more than W
    e.) Depends on whether he stands on one or two feet"

    The answer is D and I'm not sure I understand completely why, so I'm just looking for someone to check my thinking here. I think it is because the tension of the rope has an x- and y-component, and in order for the performer to be held up, the y-component must be greater than or equal to W. But that is just the y-component and because the tension also has an x-component, the overall tension has to be greater than W. Is my thinking here correct?

    The second:
    "A 12 kg crate rests on a horizontal surface and a boy pulls on it with a foce that is 30 degrees above the horizontal. If the coefficient of static friction is μs=0.52, what is the minimum force he needs to start the crate moving?"

    I drew a free body diagram and resolved the forces into x- and y-components to find:

    Along x: Fcosθ - Ff = max
    where F is the force by the boy and Ff is the friction force

    Along y: N - mg + Fsinθ = may = 0
    where N is the normal force and mg is the weight of the crate.

    So N = mg - Fsinθ
    Ffmin move = μs*N = μs*(mg - Fsinθ)

    I now have an expression I can substitute for Ff in the "Along x" equation. However, in order to solve, I have to set max = 0, in order to get:

    Ff = Fcosθ

    So, μs*(mg - Fsinθ) = Fcosθ

    And then I can solve for F. After all that rambling, my question is: how can I say the acceleration along x is 0 if I'm looking for the minimum force that will move the crate?

    "A stuntman is practicing for a 'death-defying' loop-the-loop stunt on his motorcycle. The track is essentially a circle that is perfectly vertical, and he is riding his motorcycle inside the circle. The circular loop has a radious of 3.0 m. Together, the mass of the stuntman and his motorcycle is 200 kg. If he drives his motorcycle very fast, he is sure to stay on the track at all times while doing the stunt. If he drives too slowly, he will not stay on the track (resulting in a crash).

    Calculate the smallest speed necessary for the stuntman and motorcycle to stay on the track."

    I've attached the diagram from the solutions (second attachment), showing the Free Body Diagram of the motorcycle at the top and bottom of the track.

    The solution says: "The motorcycle just barely stays on the track at the top if N = 0" and then proceeds to solve mg = mv2/r for v.

    I was wondering if anybody could help me to understand why:

    -The lengths of the force vectors are the way they are (N is shorter than mg at the top, mg is shorter than N at the bottom)
    -N = 0 at the top if the motorcycle just barely stays on the track.

    I did understand this about a week ago... but it was such a stretch that it kind of slipped my mind. Any help is appreciated!

    Attached Files:

  2. jcsd
  3. Nov 14, 2006 #2


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    Your reasoning appears sound to me. However, note that if the y component of the tension was greater than the weight of the trapeze artist he/she would be accelerated upwards :wink:
    There is no need to state this, it is taken for granted that the crate will be in equilibrium when the minimum force is applied. This is how it was when I took my mechanics exam, perhaps others have a different opinion?
    Think about centripetal acceleration :wink:
  4. Nov 14, 2006 #3
    Oops! I wasn't even thinking of that, my bad...

    Yeah, I think you're probably right, I'm just worried that the solutions are a bit brief compared to what might be expected on an exam. Maybe I'll just talk to my prof about it.

    Centripetal acceleration is toward the center of the circle, so then at the bottom the normal force is longer because there must be a net force toward the center of the circle. At the top... still not sure how I should think of this. At the top the normal force does not counteract the weight force. Since the motorcycle is accelerating downward, the track exerts less of a force on the motorcycle...?

    Still confused about N = 0 on the top... wouldn't that mean that the motorcycle is no longer in contact with the track?
  5. Nov 14, 2006 #4


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    Yes, so basically at the bottom the net force (toward the centre) is [itex]\sum\vec{F} = N-mg[/itex] and this must provide the centripetal acceleration, therefore N must be larger.

    At the top the net force (toward the centre) is [itex]\sum\vec{F}=N+mg[/itex] so both the normal force and the weight will be providing the centripetal acceleration.
    We say that as the normal force approaches zero, the motorcycle is about to leave the track, therefore if we find the point when the normal is zero this is the point where he will leave the track if he reduces his speed. Does that make sense?
    Last edited: Nov 14, 2006
  6. Nov 14, 2006 #5

    Doc Al

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    Staff: Mentor

    The only forces available to provide the centripetal force are weight and normal force. At the bottom, weight acts away from the center, so the normal force must be greater than the weight; at the top, weight and normal force add together to provide the centripetal force.

    N = 0 is the transition point. Any slower, and the motorcycle leaves the track.

    (Oops... Hoot took care of it. :wink: )
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