# Friction and energy dissipation

**URL:** <https://boards.straightdope.com/t/friction-and-energy-dissipation/191478>\
**Category:** Factual Questions\
**Created:** [July 28, 2003, 6:06pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478 "2003-07-28T18:06:33Z")\
**Posts on this page:** 10\
**Page:** 1

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**Author:** ![Trigonal\_Planar](https://avatars.discourse-cdn.com/v4/letter/t/0ea827/32.png) [@Trigonal\_Planar](https://boards.straightdope.com/u/Trigonal_Planar)\
**Post date:** [July 28, 2003, 6:06pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/1 "2003-07-28T18:06:33Z")

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When pushing on an object so as to make it move, some of the energy you supply goes into moving the object, while some of it is used up as heat in overcoming kinetic friction.

Now, what if the objection doesn’t move? You push on it, but not quite with enough force to overcome static friction. In this case, how is the energy dissipated?

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**Author:** ![Turek](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/turek/32/18782_2.png) [@Turek](https://boards.straightdope.com/u/Turek)\
**Post date:** [July 28, 2003, 6:16pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/2 "2003-07-28T18:16:01Z")

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WAG… Heat, on the surface of the object you’re pushing where it contacts the surface it’s sitting on.

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**Author:** ![Trigonal\_Planar](https://avatars.discourse-cdn.com/v4/letter/t/0ea827/32.png) [@Trigonal\_Planar](https://boards.straightdope.com/u/Trigonal_Planar)\
**Post date:** [July 28, 2003, 6:37pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/3 "2003-07-28T18:37:34Z")

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I don’t see how that could work…there’s nothing moving to generate heat. Taken to its logical extreme: Suppose I push a brick against a cement wall…would side of contact get hot just from me pushing it?

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**Author:** ![Grey](https://avatars.discourse-cdn.com/v4/letter/g/b782af/32.png) [@Grey](https://boards.straightdope.com/u/Grey)\
**Post date:** [July 28, 2003, 6:54pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/4 "2003-07-28T18:54:16Z")

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No movement by the object means no work done on the object.

Now that’s not to say that there was no work done _within_ your body.

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**Author:** ![Turek](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/turek/32/18782_2.png) [@Turek](https://boards.straightdope.com/u/Turek)\
**Post date:** [July 28, 2003, 6:55pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/5 "2003-07-28T18:55:51Z")

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But there IS stress on the molecules that are pressing against each other and generating the friction in the first place.

If you’re pushing a brick against a wall (straight on), then there would be no lateral component of your force, so no friction would be created.

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**Author:** ![NinetyWt](https://avatars.discourse-cdn.com/v4/letter/n/cc9497/32.png) [@NinetyWt](https://boards.straightdope.com/u/NinetyWt)\
**Post date:** [July 28, 2003, 7:35pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/6 "2003-07-28T19:35:43Z")

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Remember “for every action there is an equal and opposite reaction”. In the case of pushing a brick against a wall, the force you push with pushes back on you. This is evident when your hand begins to hurt. 😃

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**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\
**Post date:** [July 28, 2003, 7:47pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/7 "2003-07-28T19:47:38Z")

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> [@](#):
>
> \*Originally posted by Trigonal Planar \*  
> Now, what if the objection doesn’t move? You push on it, but not quite with enough force to overcome static friction. In this case, how is the energy dissipated?

As **Grey** said, this creates no work, and it doesn’t take any energy to do, so there’s no energy to dissipate.

Imagine if you had a book leaning against a box. The book provides a force on the box, but not enough to overcome static friction. Since the system is in a state of equilibrium, any work done would be free energy!

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**Author:** ![Trigonal\_Planar](https://avatars.discourse-cdn.com/v4/letter/t/0ea827/32.png) [@Trigonal\_Planar](https://boards.straightdope.com/u/Trigonal_Planar)\
**Post date:** [July 28, 2003, 7:48pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/8 "2003-07-28T19:48:03Z")

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Yes, true enough.

So lets look at my original example. I’ve seen graphs of energy dissipation in overcoming static friction and maintaining movement over kinetic friction. Its a curved line that reaches a maximum, then drops down to an above-zero value. As I’m increasing my force on an object in order to overcome static friction, it is not moving during that time. Its only when I exert enough force greater then static friction that the object moves. So where is this energy going whilst I’m “building up” the force?

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**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\
**Post date:** [July 28, 2003, 9:12pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/9 "2003-07-28T21:12:45Z")

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What exactly do the axes of these graphs say? “Energy disspiation” and what?

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**Author:** ![Giraffe](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giraffe/32/129_2.png) [@Giraffe](https://boards.straightdope.com/u/Giraffe)\
**Post date:** [July 28, 2003, 9:13pm UTC](https://boards.straightdope.com/t/friction-and-energy-dissipation/191478/10 "2003-07-28T21:13:41Z")

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**Trigonal** , the energy is going into breaking the attractive interaction between the two surfaces. Imagine the atoms are held together by springs which have a release valve that detaches the spring when you stretch it past a certain point. Before you pull it far enough to break it free, you’re just building up potential energy by stretching the spring. Once you break the spring, all that energy is converted to kinetic energy. If you stop before you break it, the stored potential energy you’ve added to the system will dissipate and the system will return to equilibrium.
