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

 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 $x$ ):

${F}_{\text{net}x}={T}_{\text{L}x}-{T}_{\text{R}x}.$

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

$\begin{array}{lll}{F}_{\text{net}x}=0& =& {T}_{\text{L}x}-{T}_{\text{R}x}\\ {T}_{\text{L}x}& =& {T}_{\text{R}x}.\end{array}$

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

$\begin{array}{lll}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{L}x}}{{T}_{\text{L}}}\\ {T}_{\text{L}x}& =& {T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)\\ \text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{R}x}}{{T}_{\text{R}}}\\ {T}_{\text{R}x}& =& {T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).\end{array}$

Equating ${T}_{\text{L}x}$ and ${T}_{\text{R}x}$ :

${T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)={T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).$

Thus,

${T}_{\text{L}}={T}_{\text{R}}=T,$

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] ,

${F}_{\text{net}y}={T}_{\text{L}y}+{T}_{\text{R}y}-w=0.$

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, ${T}_{\text{L}}={T}_{\text{R}}=T$ :

$\begin{array}{lll}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{L}y}}{{T}_{\text{L}}}\\ {T}_{\text{L}y}={T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)\\ \text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{R}y}}{{T}_{\text{R}}}\\ {T}_{\text{R}y}={T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).\end{array}$

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:

$\begin{array}{lll}{F}_{\text{net}y}& =& {T}_{\text{L}y}+{T}_{\text{R}y}-w=0\\ {F}_{\text{net}y}& =& T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)+T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)-w=0\\ 2\phantom{\rule{0.25em}{0ex}}T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)-w& =& 0\\ 2\phantom{\rule{0.25em}{0ex}}T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& w\end{array}$

and

$T=\frac{w}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)}=\frac{\text{mg}}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)},$

so that

$T=\frac{\left(\text{70}\text{.}\text{0 kg}\right)\left(9\text{.}{\text{80 m/s}}^{2}\right)}{2\left(0\text{.}\text{0872}\right)},$

and the tension is

$T=\text{3900 N}.$

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:

$T=\frac{w}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}.$

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:

$T=\frac{{F}_{\perp }}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}.$

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 $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =0$ ). (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:

${w}_{\parallel }=w\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)$
${w}_{\perp }=w\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\theta \right).$

• 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

Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (a) What is magnitude of the acceleration of the two teams? (b) What is the tension in the section of rope between the teams?

1. $0.{\text{11 m/s}}^{2}$
2. $1\text{.}2×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N}$

(a) Calculate the tension in a vertical strand of spider web if a spider of mass $8\text{.}\text{00}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{kg}$ hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in [link] . The strand sags at an angle of $\text{12º}$ below the horizontal. Compare this with the tension in the vertical strand (find their ratio).

(a) $7\text{.}\text{84}×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}\text{N}$

(b) $1\text{.}\text{89}×{\text{10}}^{–3}\phantom{\rule{0.25em}{0ex}}\text{N}$ . This is 2.41 times the tension in the vertical strand.

Suppose a 60.0-kg gymnast climbs a rope. (a) What is the tension in the rope if he climbs at a constant speed? (b) What is the tension in the rope if he accelerates upward at a rate of $1\text{.}{\text{50 m/s}}^{2}$ ?

Consider the baby being weighed in [link] . (a) What is the mass of the child and basket if a scale reading of 55 N is observed? (b) What is the tension ${T}_{1}$ in the cord attaching the baby to the scale? (c) What is the tension ${T}_{2}$ in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible.

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