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Problem Help

  1. Aug 10, 2003 #1
    hi can you help me with these two exercises plz

    question 1
    a 1490 kg car is moving with a steady speed of 88km/h at a constant height h above the ground on a slop track with an inclination angle to the horizontal. if there is no side force on the tyres determine(a)the radius of curvature r of the track, (b) the minimum coefficient of friction vetween the tyres and the track for the car to move at the same height ut at a speed of 160km/h

    question 2

    a train (A) with a mass of 6500 kg is travelling at 45km/h when it strikes a second truck () which has a mass of 1800 kg and is at rest. if the coefficient of restitution for the collision is 0.69 and all resistance to motion are negligible calculate,(a) the speed of the trains A and B immediatly after the impact,(b) as a result of damage during the collision truck B has a constant friction resistance to the motion of 590 N. truck B travels along the horizontal for 8m after which the track is inclicned at a 5 degree abve the horizontal. calculate the distance the truck B will move up this incline before coming to rest. (c) determine whether the truck will remain at rest ot run down the track again.

    Thanks a lot
  2. jcsd
  3. Aug 10, 2003 #2


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    It is a policy of this forum that you must show us what you have already tried yourself- even if it is only showing us that you have looked up the definitions. Show us what you have already tried and I am sure you will get a lot of assistance.

    Looks like an interesting problem!
  4. Aug 11, 2003 #3

    for question 2,i could do the first question (a), by using the momentum conservation and then using the formula of restitution
    but when i get to part b) i get stuck....i used formula of kinetic energy and work and did T1-U(1->2)=T3 to find x which is the distance, then for c) i cant figure it out, to find if it slips or not, wot procedure should do?

    for exercise 1 i'm already stuck at (a) i cant figure it out cos i did have such an example in the textbook


    take care
  5. Aug 11, 2003 #4
    Let's start witht the first one. Since you did not give us a concrete value for h one can only solve for r with h as a variable.

    The centripetal force will be supplied by the horizontal component of the normal force between the car and the track. The general equation for centripetal force that is needed to keep the car on a specific track equals:

    Fc = (mv^2)/r

    By drawing a diagram you can find the horizontal component of the normal force starting from the force of gravity on the car. This horizontal component will be equal in value to Fc

    and will be a function of sin, cos, tan...whichever you find more convenient times the force of gravity which equals mg and of course h will be in there since the angle of the sin, cos, or tan depends on h.

    Substitute this equation...for the horizontal component of the normal force acting on the car...for Fc

    Basically set Fnx = Fc and solve for r

    and there you go.
    Last edited: Aug 11, 2003

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