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Inclined Planes and Force Components

The following method generally works for force problems in two dimensions:

  2. Draw all the forces on the object(s) in question (see Section 3.D).
  3. Decide the orientation of the axes ("horizontal" and "vertical").
  4. Divide the forces into components if necessary.
  5. Solve (Fnet)y = may and (Fnet)x = max.

For the last step, note that we often have ay = 0, leading to (Fnet)y = 0. Also, be on the lookout for the words "constant velocity" or the equivalent, since that implies ax = 0 and ay = 0, a force balance.
The principle of the independence of the components of the F = ma equation is valid even when the axes are tilted, as the next example will show.

A toy car of mass 40 grams is released at the top of an incline of plastic track, inclined 30˚ from the horizontal. The car starts from rest and travels 4 m to the floor. Assuming there is no friction, how much time does it take the car to reach the floor?



First we DRAW A DIAGRAM (Figure 5-3). In addition to gravity, the track touches the car and exerts a normal force. Since the track is inclined, the normal force points not up but perpendicular to the surface. There is no frictional force.

..\art 5 jpg\figure 5-ta.bmp
Figure 5-3

We will call "horizontal" the direction along the track and "vertical" the direction perpendicular to the track (Figure 5-3).

Next, we divide the gravitational force into components (Figure 5-4). Note that Fx and Fy are not new forces but pieces of Fgrav. If we add Fx and Fy together like vectors (tip-to-tail), then we get Fgrav. It may not be obvious that the two angles shown in Figure 5-4 should both be θ = 30˚.

Note that both angles labeled θ are complementary to the angle between Fx and Fgrav. On the other hand, in physics it is generally true that two small angles which look congruent are congruent(!).


..\art 5 jpg\figure 5-tb.bmp
Figure 5-4

Now if we look at the triangle in Figure 5-4, we can write

We care about the horizontal motion, so we write


(Fnet)x = max

The only horizontal force is Fx, so we write
Fx = max


mg sin θ = max


g sin θ = ax


ax = (10 m/s2)sin 30 = 5 m/s2

We have v1x = 0 m/s and Δx = 4 m, so we can find Δt by writing




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