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⚙️ ap physics c: mech

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🚀 Unit 2

  •  ⏱️4 min read

2.3 Newton's Laws of Motion: Third Law 🐴

Gerardo Rafael Bote

gerardo rafael bote

Caroline Koffke

caroline koffke

(editor)

⏱️ November 5, 2020

📅

Newton's Third Law is one of the most quoted lines in physics, yet sometimes it feels like the most abstract of the three laws.

Newton's Third Law of Motion: For every action, there is an equal and opposite reaction.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.52-DF9DKU3lIXhP.png?alt=media&token=38cdef25-30a5-400b-88ac-5ad54c2f9e1a

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-lJv2EPbSdQ3W.png?alt=media&token=47d8d421-7d4a-4534-8c79-35030ce9482f

Image from Wikimedia Commons

What you should take away from this law is that force pairs exist in systems.

Force Pairs are always:

  • Inside the system of the two objects that are interacting
  • The same type of force (Ex: Tension)
  • Equal magnitudes, opposite directions

If you struggle to recognize force pairs, try to circle them together or make them the same color when you draw your free body diagrams! Additionally, you should be prepared for many systems of equations!

Atwood Machines

Atwood machines are seen all the time in physics problems (AP loves them!), and they are a prime example of Newton's Third Law!

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-yjTo7ePesAwa.png?alt=media&token=a07576b8-4220-4f91-a5cc-fa15a19d316f

Image from Wikimedia Commons

By combining Newton's second and third law, you can derive some formulas to describe this scenario. First, you can usually assume the string is weightless and frictionless and that the pulley has no moment of inertia (you'll learn about this in a future unit!). This means that the only forces acting on the objects are tension and gravity.

Usually, in Atwood machine questions, you are asked to determine the acceleration of the system, so let's try to find it! For this specific instance of an Atwood machine, there are only forces in the Y direction, so let's isolate the forces acting on Mass 1 and Mass 2.

Start off by drawing Free Body Diagrams for Mass 1 and Mass 2 then writing the corresponding equations.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.52-3KV1M9DGDvxH.png?alt=media&token=175b0093-acdb-4506-b59b-18308a80e985

Now we can get rid of T (Tension) by combining these equations.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.52-WYoIK0p1uNeT.png?alt=media&token=01ebedf2-d59e-4c9f-8cfc-a2d83e083f4b

Factor out g and a:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.52-XQ0YFV8Mjtry.png?alt=media&token=fed0aefe-4f0d-423d-917e-fa193db0137e

One final step to isolate a:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-10%20at%2011.53-TNG5y7RZ9YgA.png?alt=media&token=e99e0c74-2b01-4b0f-a123-681bd22867b4

⚠️Is that it? It seems so easy... What can I get confused about with Newton's 3rd Law? TBH, a lot! There are many mistakes you can make with Newton's 3rd Law. You should take note of the following:

  • "Forces occur only in pairs. Any force by one object on another cannot exist without the existence of an equal and opposite reaction force. A single force without a reaction counterpart does not exist in the world of physics." (Paul G. Hewitt)

  • Gravity (i.e., gravitational force) is also considered with Newton's 3rd Law! As you push your weight down on Earth, the planet will also push you back up.

  • Force pairs DO NOT cancel each other out. Instead, the action force and reaction force just apply themselves to different systems. For example, take a look at the picture below.

    https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-LK8NJg1OvpJr.png?alt=media&token=6cada231-7e30-42f8-b132-71dce137e968

    Image from OpenStax

    You can probably identify that the action-reaction pair is the forces dealing with the swimmer's push on the wall. Based on the circled system, F_{wall space on feet} is considered in the swimmer's free body diagram, including her weight and the buoyancy force she experiences while swimming.

Practice Questions

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-UtMPIA8qymGp.png?alt=media&token=48266c2b-962d-4c06-b7a6-b21e6b7539c2

Image from collegeboard.org

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-pAfBZ8UTnE0P.png?alt=media&token=81942d58-5bc7-45cb-88ef-b05a5d09f87f

Image from collegeboard.org

The correct answer is D. It was obvious that the weight vector (W) is pointing down, but now the question is: where do the tensions of each side of the rope point toward? Since we want "to remain at rest and in equilibrium," it should be concluded that Σ F =0 and that there is no acceleration taking place. Therefore, we should try to make the forces in the free body diagram relatively equal to each other (i.e, make sure that each dimension in the FBD cancels out). This instantly eliminates B & C as the vectors do not cancel each other out (i.e., there is a net force taking place). Finally, it doesn't make sense that either only T_L or T_R does not carry all the weight of the child on the swing. Besides, T_L allows the child to hold the rope to make himself in static equilibrium. This eliminates A & E. Since T_L and T_R are smaller than W and equal in size, the likely answer is D.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-aH3i1ibeQ4m9.png?alt=media&token=13c2518f-0a79-4c87-9a35-2743c213bd6d

Answer

To complete this problem, you need to fully integrate Newton's 3rd Law into the relationships among the blocks. For me, I see the following relationships (i.e., net force equations for each block):

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-11%20at%2012.00-KKEajxVRI7HZ.png?alt=media&token=d4a70cb2-9ca8-49ee-bfae-5acc36b4f392

Since you know that both blocks are moving to the right with an acceleration of 2\space m/s^2, it can be declared that:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-11%20at%2012.00-yzSojOqkbaSk.png?alt=media&token=1d9c07f6-e1bf-4d60-93c6-aabccfa3a96c

Knowing Newton's 3rd Law, you can also determine that:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-11%20at%2012.00-vIAvpoj8Zqvl.png?alt=media&token=4288ebea-291f-4ce4-a66b-f972e40d4460

⚠️Wait, isn't there supposed to be a negative sign in the equation above? Yes, but that is only showing the relationship in action-reaction pairs that one force goes to the opposite way of the other force. Here, I'm showing that the action-reaction pairs are equal in magnitude.

Now you treat the following equations above by doing substitutions:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-11%20at%2012.01-mtLpDmRsmB9T.png?alt=media&token=028f4bb9-276d-4530-81b0-a78746c11ab7

Lastly, you put 28 N in the first equation above to figure the magnitude of F_{L on K}:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-11%20at%2012.01-i7czLOwAq9dR.png?alt=media&token=e31993d9-70eb-4e45-b61a-4711d795f331

If you need more help, watch this YouTube video  made by AbrahamPhysics

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-OZgrG1nvczQP.png?alt=media&token=f9c68cc6-3542-498e-a999-92c0713ffcbb

Image from Varsity Tutors

(Hint: Use g=9.81 m/s2g=9.81 m/s2)

Answer

Since Dave weighs 589 N, that must mean his mass is:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-11%20at%2012.04-vu5YaxLRvesg.png?alt=media&token=0745b42e-6c3d-4096-8a3c-b8a3a1bc08a7

Since he accelerated at  4 \space m/s^2, we can determine that he was pushed off by the ground with:

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreen%20Shot%202020-09-11%20at%2012.04-IXc9kG5NC3o9.png?alt=media&token=02276a75-6648-41a5-a900-c54f27df429a

By Newton's 3rd law of motion, Dave must also have pushed of the ground with a force of 240 Newtons.

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