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0.4 3.5 newton’s third law of motion  (Page 2/4)

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Getting up to speed: choosing the correct system

A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in [link] . Her mass is 65.0 kg, the cart’s is 12.0 kg, and the equipment’s is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart’s wheels and air resistance, total 24.0 N.

A professor is pushing a cart of demonstration equipment. Two systems are labeled in the figure. System one includes both the professor and cart, and system two only has the cart. She is exerting some force F sub prof toward the right, shown by a vector arrow, and the cart is also pushing her with the same magnitude of force directed toward the left, shown by a vector F sub cart, having same length as F sub prof. The friction force small f is shown by a vector arrow pointing left acting between the wheels of the cart and the floor. The professor is pushing the floor with her feet with a force F sub foot toward the left, shown by a vector arrow. The floor is pushing her feet with a force that has the same magnitude, F sub floor, shown by a vector arrow pointing right that has the same length as the vector F sub foot. A free-body diagram is also shown. For system one, friction force acting toward the left is shown by a vector arrow having a small length, and the force F sub floor is acting toward the right, shown by a vector arrow larger than the length of vector f. In system two, friction force represented by a short vector small f acts toward the left and another vector F sub prof is represented by a vector arrow toward the right. F sub prof is longer than small f.
A professor pushes a cart of demonstration equipment. The lengths of the arrows are proportional to the magnitudes of the forces (except for f size 12{f} {} , since it is too small to draw to scale). Different questions are asked in each example; thus, the system of interest must be defined differently for each. System 1 is appropriate for [link] , since it asks for the acceleration of the entire group of objects. Only F floor size 12{F rSub { size 8{"floor"} } } {} and f size 12{f} {} are external forces acting on System 1 along the line of motion. All other forces either cancel or act on the outside world. System 2 is chosen for this example so that F prof size 12{F rSub { size 8{"prof"} } } {} will be an external force and enter into Newton’s second law. Note that the free-body diagrams, which allow us to apply Newton’s second law, vary with the system chosen.

Strategy

Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in [link] . The professor pushes backward with a force F foot size 12{F rSub { size 8{"foot"} } } {} of 150 N. According to Newton’s third law, the floor exerts a forward reaction force F floor size 12{F rSub { size 8{"floor"} } } {} of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction. The problem is therefore one-dimensional along the horizontal direction. As noted, f size 12{f} {} opposes the motion and is thus in the opposite direction of F floor size 12{F rSub { size 8{"floor"} } } {} . Note that we do not include the forces F prof size 12{F rSub { size 8{"prof"} } } {} or F cart size 12{F rSub { size 8{"cart"} } } {} because these are internal forces, and we do not include F foot size 12{F rSub { size 8{"foot"} } } {} because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use Newton’s second law to find the acceleration as requested. See the free-body diagram in the figure.

Solution

Newton’s second law is given by

a = F net m size 12{a = { {F rSub { size 8{"net"} } } over {m} } } {} .

The net external force on System 1 is deduced from [link] and the discussion above to be

F net = F floor − f = 150 N − 24 . 0 N = 126 N size 12{F rSub { size 8{"net"} } = F rSub { size 8{"floor"} } -f ="150 N"-"24" "." "0 N"="126 N"} {} .

The mass of System 1 is

m = ( 65 . 0 + 12 . 0 + 7 . 0 ) kg = 84 kg size 12{m = \( "65" "." "0 "+" 12" "." "0 "+" 7" "." 0 \) " kg "=" 84 kg"} {} .

These values of F net size 12{F} {} and m size 12{m} {} produce an acceleration of

a = F net m , a = 1 26 N 84 kg = 1 . 5 m/s 2 . alignl { stack { size 12{a= { {F rSub { size 8{"net"} } } over {m} } ,} {} #a = { {1"26 N"} over {"84"" kg"} } =" 1" "." "5 m/s" rSup { size 8{2} } "." {} } } {}

Discussion

None of the forces between components of System 1, such as between the professor’s hands and the cart, contribute to the net external force because they are internal to System 1. Another way to look at this is to note that forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite force back on her. In this case both forces act on the same system and, therefore, cancel. Thus internal forces (between components of a system) cancel. Choosing System 1 was crucial to solving this problem.

Force on the cart—choosing a new system

Calculate the force the professor exerts on the cart in [link] using data from the previous example if needed.

Strategy

If we now define the system of interest to be the cart plus equipment (System 2 in [link] ), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, F prof size 12{F rSub { size 8{"prof"} } } {} , is an external force acting on System 2. F prof size 12{F rSub { size 8{"prof"} } } {} was internal to System 1, but it is external to System 2 and will enter Newton’s second law for System 2.

Solution

Newton’s second law can be used to find F prof size 12{F rSub { size 8{"prof"} } } {} . Starting with

a = F net m size 12{a = { {F rSub { size 8{"net"} } } over {m} } } {}

and noting that the magnitude of the net external force on System 2 is

F net = F prof − f size 12{F rSub { size 8{"net"} } = F rSub { size 8{"prof"} } -f} {} ,

we solve for F prof size 12{F rSub { size 8{"prof"} } } {} , the desired quantity:

F prof = F net + f . size 12{F rSub { size 8{"prof"} } = F rSub { size 8{"net"} } + f} {}

The value of f size 12{f} {} is given, so we must calculate net F net size 12{F} {} . That can be done since both the acceleration and mass of System 2 are known. Using Newton’s second law we see that

F net = ma size 12{F rSub { size 8{"net"} } = ital "ma"} {} ,

where the mass of System 2 is 19.0 kg ( m size 12{m} {} = 12.0 kg + 7.0 kg) and its acceleration was found to be a = 1.5 m/s 2 size 12{a=1 "." "50"" m/s" rSup { size 8{2} } } {} in the previous example. Thus,

F net = ma size 12{F rSub { size 8{"net"} } = ital "ma"} {} ,

F net = ( 19 . 0 kg ) ( 1.5 m/s 2 ) = 29 N size 12{F rSub { size 8{"net"} } = \( "19" "." "0 kg" \) \( 1 "." "50 m/s" rSup { size 8{2} } \) ="28" "." 5" N"} {} .

Now we can find the desired force:

F prof = F net + f size 12{F rSub { size 8{"prof"} } =F rSub { size 8{"net"} } +f} {} ,

F prof = 29 N + 24.0 N = 53 N size 12{F rSub { size 8{"prof"} } ="28" "." 5" N "+"24" "." "0 N "="52" "." "5 N"} {} .

Discussion

It is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor.

The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which is not necessarily the same thing).

Section summary

  • Newton’s third law of motion    represents a basic symmetry in nature. It states: Whenever 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 the first body exerts.
  • A thrust    is a reaction force that pushes a body forward in response to a backward force. Rockets, airplanes, and cars are pushed forward by a thrust reaction force.

Conceptual questions

When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat—is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.)

An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will experience an equal and opposite force from the other player. Use Newton’s laws and draw a free-body diagram of an appropriate system to explain how he can still out-push the opposition if he is strong enough.

Newton’s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the “system of interest” affects whether one such pair of forces cancels.

Problem exercises

What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2 . 40 × 10 4 m/s 2 size 12{2 "." "40"´"10" rSup { size 8{4} } " m/s" rSup { size 8{2} } } {} ? What is the magnitude of the force exerted on the ship by the artillery shell?

Force on shell: 2 . 64 × 10 7 N size 12{2 "." "64" times "10" rSup { size 8{7} } `N} {}

Force exerted on ship = − 2 . 64 × 10 7 N size 12{ - 2 "." "64" times "10" rSup { size 8{7} } `N} {} , by Newton’s third law

A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1 . 20 m/s 2 size 12{1 "." "20"" m/s" rSup { size 8{2} } } {} backward. (a) What is the force of friction between the losing player’s feet and the grass? (b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110 kg? (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation.
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Source:  OpenStax, Newton's laws. OpenStax CNX. Oct 25, 2015 Download for free at https://legacy.cnx.org/content/col11898/1.1
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