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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 n e t = 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 n e t = 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 = m g

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

T = m g = ( 5 . 00 k g ) ( 9 . 80 m / s 2 ) = 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 n e t = F f l o o r - f = 150 N - 24 . 0 N = 126 N

    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.

Alt text
A teacher pushes a cart of demonstration equipment.

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 n e t = F f l o o r - f = 150 N - 24 . 0 N = 126 N )

Newton’s second law is a = F n e t 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 n e t = F f l o o r - f = 150 N - 24 . 0 N = 126 N

The mass of the system is the sum of the mass of the teacher, cart, and equipment: m = ( 65 . 0 + 12 . 0 + 7 . 0 ) k g = 84 k g

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

  1. a = 126 N 84 k g = 1 . 5 m / s 2
    1. F net
      1. equipment
      2. teacher
      3. cart
    2. m
    3. F floor
  2. a = F n e t m
  3. a = 126 N 84 k g = 1 . 5 m / 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.

Key equations

Normal force for a non-accelerating horizontal surface:

N = m g

Tension for an object at rest:

T = m g

Check your understanding

Concept items

Performance task

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

what is math number
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Need help solving this problem (2/7)^-2
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x+2y-z=7
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-1
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lim x to infinity e^1-e^-1/log(1+x)
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If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
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make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
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20/(×-6^2)
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A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
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Source:  OpenStax, Updated tutor hs physics content - legacy. OpenStax CNX. Mar 16, 2015 Download for free at https://legacy.cnx.org/content/col11768/1.4
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