Solution to the inclined plane problems given at Newton.htm

1. No friction case:
Forces on M are FN, Mg, and tension T. We need components parallel and perpendicular to the plane, so Mg must be resolved into its components, Mgsinq and Mgcosq , where M= 2.00 kg, g = 9.8 N/kg, and q = 20.0o. Mgsinq is in the downhill direction, parallel to the plane, and this is less than mg (recall m = 1.20 kg), so it will accelerate uphill.
For each mass, let the positive direction be the direction of the acceleration, and write the sum of forces in the acceleration direction = ma for each mass.
T - Mgsinq = Ma
mg - T = ma
Add these together and the T disappears, and we find that
mg - Mgsinq = (M+m)a
Plug in the numbers and get a = 1.58 m/s2.                                                        20o

Another way to do the above: Lump M+m together and pull one way with force mg and the opposite way with Mgsinq.

  1. This problem was to find the minimum m s to prevent acceleration.
    (Does your browser recognize ^ as the perpendicular symbol?)
    Note that the normal force FN or N is ^ to the plane up and to the left, and Mgcosq is opposite; these must add up to zero since there is no acceleration ^ to the plane, so FN = Mgcosq . (Oops, in the sketch I called it N, not FN.) The static friction force is m sFN, and it must be in the downhill direction to prevent acceleration. Parallel to the plane, the forces add up to zero, or
    mg - m sFN - Mgsinq = 0. Plug in FN = Mgcosq , do the numbers and find m s = 0.275.
  2. This problem was to find the acceleration if the coefficient of kinetic friction is 0.200. This is the same as part a) except that there is an additional force
    m Mgcosq in the downhill direction when the thing is sliding uphill. So use the method in part a) but include this force, and your final equation becomes
    mg- Mgsinq - m Mgcosq = (M+m)a.
    Plug and chug and get a = 0.429 m/s2.




    Here is an alternative approach which is simpler in a way. Notice that for M, T is in the direction of a and for m, T opposes a, so as a whole, T drops out. In your head or on paper, visualize the system like this, with opposing forces mg and Mgsinq :

The net force on the system is mg - Mgsinq , the total mass is M+m, and the acceleration is net force/total mass.

In 2 there is an additional force to the left to make the net force zero.

In 3 there is also another force to the left, but it is smaller than in 2, so acceleration occurs.

Now you can force this out of here and bring back the force and acceleration stuff or the main physics page.

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