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Force exerted on motorcycle and rider

  1. Jun 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A motorcycle and 60.0 kg rider accelerate at 3.0 m/s^2 up a ramp inclined 10 degrees above the horizontal. (a) What is the magnitude of the net force acting on the rider? (b) What is the magnitude of the force on the rider from the motorcycle?

    2. Relevant equations
    Net Force = Mass x Acceleration
    Answer given for first part: 180 N
    Answer given for second part: 640 N

    3. The attempt at a solution
    The first part is easily solved: Since the mass is given to be 60 kg, and the acceleration up the incline is 3 m/s^2, therefore the net force is 60 x 3 = 180 N. This matches the answer given.

    For the second part, I thought it could be done by taking the normal force acting on the motorcycle, which on the given incline, would be = mg cos 10 degrees = 60 x 9.8 x cos 10 degrees = 579 N.

    This is way off the answer given, which is 640 N. Did I get some theoretical bit wrong? I really don't see any other force which should be acting - after all, the mass of the motorcycle is not given.
     
  2. jcsd
  3. Jun 17, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi ankurb,

    What are all the forces acting on the rider? There has to be more than just that normal force you calculated and the weight because neither of those has a component up the incline, yet the rider is accelerating up the incline.
     
  4. Jun 19, 2008 #3
    Solved

    Oh. Right! I thought we were only supposed to consider the vertical force due to normal reaction. If I factor in the force exerted by the motorcycle to go up the incline, then I get the answer.

    -mg sin theta + F = ma
    or, F = m(g sin theta + a) = 60 x (9.8 sin 10 deg + 3) = 282 N

    Thus, total force exerted or rider = ( 282^2 + 579^2 ) ^ 1/2 = 640 N (approx).

    Thanks! :)
     
  5. Mar 16, 2009 #4
    hello,

    i'm working on this same problem and am having the same problem with it as the thread starter, so i hope no one minds if i revive this old thread. for part b) i initially got 579 N as well, but when i read alphysicist's reply, i went back and ended up with 584 N. i know what i am doing wrong -- for the combined horizontal forces, i'm assuming that mg sin theta and ma should be subtracted from each other, rather than added -- but i'm having trouble seeing why this wouldn't be the case. why aren't those two horizontal forces going in opposite directions?

    hope someone can help. thanks.
     
  6. Oct 4, 2011 #5
    I have the same problem too, and I don't really understand it. This is how I approached the problem. I first took into account the mass of the Motorcycle and the mass of the rider as well. So I tried to do FBD on each of them

    Motorcycle
    -m1gsin 10 + F = m1a
    Rider
    -mgsin10 = ma

    Now I have no idea to solve it, since m1 is not given and I can't figure out a way to find it either. So basically I think that I thought was wrong. But I don't understand is, the problem did mention the motorcycle and it didn't say anything about ignore the mass it its, so what is the reason that we should ignore it ? I mean, in the future, if I come across another problem like this, how should I know when mass of the motorcycle is going to be ignored or not ?
     
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