# 1.4 Newton's third law of motion  (Page 4/8)

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Tension in the rope must equal the weight of the supported mass, as we can prove by using Newton’s second law. If the 5.00 kg mass in the figure is stationary, then its acceleration is zero, so ${F}_{net}=0$ . The only external forces acting on the mass are its weight W and the tension T supplied by the rope. Summing the external forces to find the net force, we obtain

${F}_{net}=T-W=0$
where T and W are the magnitudes of the tension and weight and their signs indicate direction, with up being positive. By substituting mg for Fnet and rearranging the equation, the tension equals the weight of the supported mass, just as you would expect:
$T=W=mg$

For a 5.00 kg mass (neglecting the mass of the rope), we see that

$T=mg=\left(5.00 kg\right)\left(9.80 m/{s}^{2}\right)=49.0 N$

Another example of Newton’s third law in action is thrust. Rockets move forward by expelling gas backward at high velocity. This means that the rocket exerts a large force backward on the gas in the rocket combustion chamber, and the gas in turn exerts a large force forward on the rocket in response. This reaction force is called thrust.

A common misconception is that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can expel exhaust gases more easily.

## Newton’s third law of motion

This video explains Newton’s third law of motion through examples involving push, normal force, and thrust (the force that propels a rocket or a jet). Think about this equation as you watch: ${F}_{net}={F}_{floor}-f=150N-24.0N=126N$

## Words defined in the video:

1. Force
2. Push
3. Normal force
4. Thrust

A physics teacher pushes a cart of demonstration equipment to a classroom, as in Image 4.12 Her mass is 65.0 kg, the cart’s mass is 12.0 kg, and the equipment’s mass is 7.0 kg. To push the cart forward, the teacher’s foot applies a force F foot of 150 N in the opposite direction (backward) on the floor. Calculate the acceleration produced by the teacher. The force of friction, which opposes the motion, is 24.0 N.

We should not include the forces F teacher , F cart , or F foot because these are exerted by the system, not on the system. We find the net external force by adding together the external forces acting on the system (see free-body diagram in the figure) and then use Newton’s second law to find the acceleration.

## Forces that shouldn’t be included:

1. F teacher (m/s 2 )
2. F cart
3. F foot ( ${F}_{net}={F}_{floor}-f=150N-24.0N=126N$ )

Newton’s second law is $a=\frac{{F}_{net}}{m}$

The net external force on the system is the sum of the external forces: the force of the floor acting on the teacher, cart, and equipment (in the horizontal direction) and the force of friction. Because friction acts in the opposite direction, we assign it a negative value. Thus, for the net force, we obtain ${F}_{net}={F}_{floor}-f=150N-24.0N=126N$

The mass of the system is the sum of the mass of the teacher, cart, and equipment: $m=\left(65.0+12.0+7.0\right)kg=84kg$

Insert these values of net F and m into Newton’s second law to obtain the acceleration of the system:

1. $a=\frac{126N}{84kg}=1.5m/{s}^{2}$
1. F net
1. equipment
2. teacher
3. cart
2. m
3. F floor
2. $a=\frac{{F}_{net}}{m}$
3. $a=\frac{126N}{84kg}=1.5m/{s}^{2}$

## Section summary

• Newton’s third law of motion states that, when one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that it exerts.
• When objects rest on a surface, the surface applies a force on the object that opposes the weight of the object. This force acts perpendicular to the surface and is called the normal force.
• The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension. When a rope supports the weight of an object at rest, the tension in the rope is equal to the weight of the object.
• Thrust is a force that pushes an object forward in response to the backward ejection of mass by the object. Rockets and airplanes are pushed forward by thrust.

## Normal force for a non-accelerating horizontal surface:

$N=mg$

## Tension for an object at rest:

$T=mg$

## Concept items

To investigate how mass affects tension in a connector, find a rubber band and some objects to hang from the end of the rubber band. How much does the rubber band stretch when a light object is hung from it?How much does it stretch when a heavier object is suspended? What does this show?Measure the mass of the object and the corresponding length of the rubber band in each case and plot a graph of mass vs length. (15 min) [link]

#### Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
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20/(×-6^2)
Salomon
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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