# 3.9 Normal, tension, and other examples of forces  (Page 7/10)

 Page 7 / 10

All the forces discussed in this section are real forces, but there are a number of other real forces, such as lift and thrust, that are not discussed in this section. They are more specialized, and it is not necessary to discuss every type of force. It is natural, however, to ask where the basic simplicity we seek to find in physics is in the long list of forces. Are some more basic than others? Are some different manifestations of the same underlying force? The answer to both questions is yes, as will be seen in the next (extended) section and in the treatment of modern physics later in the text.

## Phet explorations: forces in 1 dimension

Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

## 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:

$T=\text{mg}.$

• In any inertial frame of reference (one that is not accelerated or rotated), Newton’s laws have the simple forms given in this chapter and all forces are real forces having a physical origin.

## Conceptual questions

If a leg is suspended by a traction setup as shown in [link] , what is the tension in the rope?

In a traction setup for a broken bone, with pulleys and rope available, how might we be able to increase the force along the tibia using the same weight? (See [link] .) (Note that the tibia is the shin bone shown in this image.)

## 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. $1\text{.}2×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N}$

What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at $7\text{.}{\text{50 m/s}}^{2}$ ? Note that the answer is independent of the velocity of the gymnast—she can be moving either up or down, or be stationary.

(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 $1\text{.}{\text{50 m/s}}^{2}$ ?

Show that, as stated in the text, a force ${\mathbf{\text{F}}}_{\perp }$ exerted on a flexible medium at its center and perpendicular to its length (such as on the tightrope wire in [link] ) gives rise to a tension of magnitude $T=\frac{{F}_{\perp }}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}$ .

Newton’s second law applied in vertical direction gives

${F}_{y}=F-2T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =0$
${F}_{}=2T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$
$T=\frac{{F}_{}}{\text{2 sin}\phantom{\rule{0.25em}{0ex}}\theta }.$

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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