# Inclined Planes and Force Components

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

- DRAW A DIAGRAM.
- Draw all the forces on the object(s) in question (see Section 3.D).
- Decide the orientation of the axes ("horizontal" and "vertical").
- Divide the forces into components if necessary.
- Solve (
*F*_{net})_{y }=*ma*_{y }and (*F*_{net})_{x }=*ma*_{x}.

*a*

_{y }= 0, leading to (

*F*

_{net})

_{y }= 0. Also, be on the lookout for the words "constant velocity" or the equivalent, since that implies

*a*

_{x }= 0 and

*a*

_{y }= 0, a force balance.

*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 30Ëš 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.

**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 *F*_{x }and *F*_{y }are not new forces but pieces of *F*_{grav}. If we add *F*_{x }and *F*_{y }together like vectors (tip-to-tail), then we get *F*_{grav}. It may not be obvious that the two angles shown in Figure 5-4 should both be *Î¸* = 30Ëš.

Note that both angles labeled *Î¸* are complementary to the angle between *F*_{x }and *F*_{grav}. On the other hand, in physics it is generally true that two small angles which look congruent are congruent(!).

**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

(*F*_{net})_{x} = *ma*_{x}

The only horizontal force is *F*_{x}, so we write

*F*_{x} = *ma*_{x}

*mg* sin *Î¸* = *ma*_{x}

*g* sin *Î¸* = *a*_{x}

*a*_{x} = (10 m/s^{2})sin 30 = 5 m/s^{2}

We have *v*_{1x} = 0 m/s and Î”*x* = 4 m, so we can find Î”*t* by writing