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Coefficient of static friction of car

  1. Apr 16, 2004 #1
    Please help....

    I need help with this 2 physics questions...thanks in advance..

    1. A car can negotiate an unbanked curve safely at a certain maximum speed when the coefficient of static friction between the tires and the ground is 0.946. At what angle should the same curve be banked for the car to negotiate the curve safely at the same maximum speed without relying on friction?

    2. A satellite is in a circular earth orbit that has a radius of 7.63 x 106 m. A model airplane is flying on a 22.3-m guideline in a horizontal circle. The guideline is nearly parallel to the ground. Find the speed of the plane such that the plane and the satellite have the same centripetal acceleration.
     
  2. jcsd
  3. Apr 16, 2004 #2
    1) Whoa, a coefficient of static friction? You mean the car's not moving?!

    Anyway, find the force that the kinetic friction is causing.

    Now assume that the road is curved, so therefore the normal force will be angled. Decompose that normal force into two component forces, horizontal and vertical. The vertical normal force must support the downward force on the car (force of gravity) while the horizontal force must replace the force that the kinetic friction used to provide. So you have a triangle and you know two sides of it.

    What do you do when you have two sides of a triangle and want the angle?

    2) a_gravity = G*M_earth/r^2

    In order for the satellite to be in orbit, the centripetal force must be equal to the force of gravity (they are, in fact, one and the same). So that means that the plane must experience the same centripetal acceleration as you just calculated above. Remember the formula for centripetal acceleration?

    a = v^2/r

    So...

    GM/r_orbit = v^2/r_cord

    G is known and constant, M is known and constant, r_orbit is known and constant, r_cord is known and constant. Solve for v.

    cookiemonster
     
  4. Apr 17, 2004 #3

    Doc Al

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    Staff: Mentor

    it better be static friction!

    The friction acts sideways to the velocity of the car--it's providing the centripetal force. There's no motion of the car in that direction, so it is static friction. Otherwise the car is sliding off the road!

    But your analysis is perfectly correct. :smile:
     
  5. Apr 17, 2004 #4
    Hm... Good point.

    cookiemonster
     
  6. Apr 18, 2004 #5
    thanks for the help..but for number 2 i got an answer of 34143.15...i don't know if that's right or not, i think is wrong... but i followed your equation...so i don't know...

    for number 1 i still don't understand....
     
  7. Apr 18, 2004 #6
    #2) Have I convinced you that I'm right?

    #1) When the car drives around the curve, the ground exerts a normal force on it. This normal force is the force that keeps the car from burrowing right through the ground (because that's not going to happen in the real world, now is it?). Which direction do you suppose this normal force will be pointed? Upward and toward the inside of the circle, right? Now this isn't purely in either the vertical or horizontal directions, so what we do is decompose it into its component vectors, the vertical component and the horizontal component.

    With me so far?

    All right, so we got a horizontal component of the normal force and a vertical component of the normal force. Let's look at what other forces are acting on our car. We have gravity, obviously, since the car's on Earth. Gravity is pointed downward and has magnitude mg. Also, our car is in accelerated motion: its motion is constantly changing direction such that its path will be a circle. So we want there to be a left-over centripetal force. Which direction does a centripetal force act? Horizontally toward the center of the circle. Its magnitude? You calculated it in the first part when you looked at the static friction.

    So what do we know so far. We know we have a normal force and it has two components, a vertical a a horizontal one. We know we have the force of gravity, which is directed downward and has magnitude mg. And we know that we want to have a net force (that is, the sum of all of our forces; i.e. our leftover force) inward with magnitude the static friction force you calculated in the first part.

    So what does this mean? This means that the vertical component of our normal force and gravity must cancel each other (since there is no net vertical force, or else our car would be flying or burrowing into the ground). So what's the vertical component of our normal force?

    Additionally, the only force acting in the horizontal direction is the horizontal component of the normal force. We want a force acting in the horizontal direction, and we want it to have a certain magnitude. So that means that it must be that horizontal component of the normal force. So what's the horizontal component of the normal force gotta be?

    All right, now we know both components of the normal force. So we can make a triangle out of the normal force. We know two sides of it, so we can calculate the angle. How do you do that?

    cookiemonster
     
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