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Horizontal and Vertical Motion, Again

In the last chapter we solved problems with gravity as the only force. Well, gravity is a fine force indeed, but we need to understand problems in which other forces are present. That is the goal of this chapter.
In the last chapter we discussed the independence of vertical and horizontal motion for objects in freefall, but it turns out the principle works when other forces are acting as well:

Independence of Vertical and Horizontal Motion

An object has forces  acting on it. If  the vertical components of the forces, then we have


Similarly, if  are the horizontal components of the

forces, then we have




This is a more useful form of Fnet = ma, the equation that we discussed in Section 3.B. An example will help illustrate how the principle in the above box is used to solve problems.

In the following example, we find a toy wagon rolling on the ground. In general, when an object is on the ground or some other surface, that surface exerts one or two forces on it: always a normal force N pointing perpendicular to the surface and sometimes a frictional force Ffric pointing parallel to the surface. We will postpone discussing friction until Chapter 6.


A boy pulls a red wagon (10 kg) with a constant force 20 N. The handle makes an angle 30˚ with the horizontal. Assume there is no friction.

  1. If the wagon starts from rest, how fast is it going after 3 seconds?
  2. What is the normal force acting on the wagon?

First we DRAW A DIAGRAM (Figure 5-1) showing all the forces on the wagon body. The handle and the ground are the only two things which touch it. So in addition to gravity, there are the tension due to the handle and the normal force. The tension T points along the handle, that is, 30˚ from the horizontal.


..\art 5 jpg\figure 5-tc.bmp
Figure 5-1

In Figure 5-2 we resolve the tension into components (recall trigonometry), so we have


..\art 5 jpg\figure 5-td.bmp
Figure 5-2

The normal force and gravity do not have horizontal components, so the only horizontal force is Tx. Thus we write


(Fnet)x = max,
Tx = max,
ax, = Tx/m = 17N/10kg = 1.7 m/s2

Using this and v1x = 0 m/s and Δt = 3 s, we derive a horizontal velocity


v2x = v1x + axΔt
= 0 m/s + 1.7 m/s2(3s) = 4.1 m/s

which is the answer to part a.

If we consider the vertical components of force, then we can find (Fnet)y just by looking at the diagram, so we write


(Fnet)y N + Ty – Fgrav

where we use the positive sign for forces which point up; negative for down. Now the second law of motion connects this with vertical acceleration, so that


f(Fnet)y may

But the cart is not moving vertically, so we know that vy is constant (and zero) and thus ay is zero. This means (Fnet)y is zero, so we have


0 = N + Ty – Fgrav,
N = Fgrav – Ty
mg – Ty
= (10kg)(10m/s2) – 10N

= 90 N.

Notice that the normal force is not the same as the gravitational force, a mistake often made by students. Why is the normal force not the same as gravity?

Note that in solving part a, we used information about forces to obtain the horizontal acceleration ax and then the answer. In part b we reasoned the other way, using information about the vertical acceleration ay = 0 to obtain information about the normal force. This strategy of reasoning in both directions will be useful in many problems.

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