0.5 3.6 normal and tension forces  (Page 5/8)

Solution

First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system.

Consider the horizontal components of the forces (denoted with a subscript $x$ ):

The net external horizontal force $F_{\text{net}x}=0$ , since the person is stationary. Thus,

Now, observe [link] . You can use trigonometry to determine the magnitude of $T_{\text{L}}$ and $T_{\text{R}}$ . Notice that:

Equating $T_{\text{L}x}$ and $T_{\text{R}x}$ :

Thus,

as predicted. Now, considering the vertical components (denoted by a subscript $y$ ), we can solve for $T$ . Again, since the person is stationary, Newton’s second law implies that net $F_y=0$ . Thus, as illustrated in the free-body diagram in [link] ,

Observing [link] , we can use trigonometry to determine the relationship between $T_{\text{L}y}$ , $T_{\text{R}y}$ , and $T$ . As we determined from the analysis in the horizontal direction, $T_{\text{L}}=T_{\text{R}}=T$ :

Now, we can substitute the values for $T_{\text{L}y}$ and $T_{\text{R}y}$ , into the net force equation in the vertical direction:

and

so that

and the tension is

Discussion

Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker.

If we wish to create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in [link] . As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way:

We can extend this expression to describe the tension $T$ created when a perpendicular force ( ${\mathbf{\text{F}}}_{\perp }$ ) is exerted at the middle of a flexible connector:

Note that $\theta$ is the angle between the horizontal and the bent connector. In this case, $T$ becomes very large as $\theta$ approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., $\theta =0$ and $\text{sin}\theta =0$ ). (See [link] .)

Section summary

• When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting force acts perpendicular to and away from the surface. It is called a normal force, $\mathbf{\text{N}}$ .
• When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object:
  $N=\text{mg}.$

• When objects rest on an inclined plane that makes an angle $\theta$ with the horizontal surface, the weight of the object can be resolved into components that act perpendicular ( ${\mathbf{\text{w}}}_{\perp }$ ) and parallel ( ${\mathbf{\text{w}}}_{\parallel }$ ) to the surface of the plane. These components can be calculated using:
  $w_{\parallel }=w\text{sin}(\theta )=\text{mg}\text{sin}(\theta )$
  $w_{\perp }=w\text{cos}(\theta )=\text{mg}\text{cos}(\theta ).$

• The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, $\mathbf{\text{T}}$ . When a rope supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object:
  $T=\text{mg}.$

Problem exercises