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

...(1) | |

...(2) |

Similarly, if are the horizontal components of the

forces, then we have

...(3) | |

...(4) |

* *

*F*=

_{net}*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.

*N*pointing perpendicular to the surface and sometimes a frictional force

*F*

_{fric }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 30Ëš with the horizontal. Assume there is no friction.

- If the wagon starts from rest, how fast is it going after 3 seconds?
- 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, 30Ëš from the horizontal.

**Figure 5-1**

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

**Figure 5-2**

The normal force and gravity do not have horizontal components, so the only horizontal force is *T*_{x}. Thus we write

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

*T*_{x} = *ma*_{x},

*a*_{x}, = *T*_{x}/*m* = 17N/10kg = 1.7 m/s^{2}

Using this and *v*_{1x} = 0 m/s and Î”*t* = 3 s, we derive a horizontal velocity

*v*_{2x} = *v*_{1x} + *a*_{x}Î”*t*

= 0 m/s + 1.7 m/s^{2}(3s) = 4.1 m/s

which is the answer to part a.

If we consider the vertical components of force, then we can find (*F*_{net})_{y} just by looking at the diagram, so we write

(*F*_{net})_{y }= *N* + *T*_{y }â€“ *F*_{grav}

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

But the cart is not moving vertically, so we know that *v*_{y }is constant (and zero) and thus *a*_{y }is zero. This means (*F*_{net})_{y }is zero, so we have

0 = *N* + *T*_{y }â€“ *F*_{grav},

*N* = *F*_{grav }â€“ *T*_{y}

= *mg* â€“ *T*_{y}

= (10kg)(10m/s^{2}) â€“ 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?

*a*

_{x }and then the answer. In part b we reasoned the other way, using information about the vertical acceleration

*a*

_{y }= 0 to obtain information about the normal force. This strategy of reasoning in both directions will be useful in many problems.