# Friction and power problem

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1. Oct 31, 2014

### AlexPilk

1. The problem statement, all variables and given/known data
There's a TIR with a mass of 2000 tons. Constant velocity = 10 m/s, coefficient of friction = 0.05.
What is the power of the TIR

2. Relevant equations
P=F*v
F(friction)=uN=umg

3. The attempt at a solution
So there are 2 forces acting on the TIR:
1. The one pushing it forward which is equal = 0 since acceleration = 0. (which doesn't make any logical sense to me, but since F=ma)
2. Friction = umg = 0.05*2 000 000*10=1 000 000 N

As far as I understand the first force isn't equal to 0, because then the overall force on the TIR is 0-150000=-150000 N and the power is -1 500 000, which doesn't make any sense either

Last edited: Oct 31, 2014
2. Oct 31, 2014

### BvU

No idea what a TIR is, except in Europe it's Transports Internationaux Routiers and them trucks don't go much over 50 tons.... :)

I do hope the 10 in your expression stands for g, otherwise the dimensions don't match. The numbers certainly don't match: 0.05*2000000*10 = 1 MN and not 150 kN.

The net horizontal force on the truck (let's assume it's a truck) is zero: constant speed. There are two components: the one pushing it forward and the one holding it back (the friction -- if we forget the aerodynamic drag, which may be realistic at this low speed). Together they are 0. So the one pushing it forward is -umg. Your relevant formula comes up with a huge power, even for this very moderate speed.

Once again you run into an exercise that doesn't feel good. Is it your book, your teacher, or something else ?

3. Oct 31, 2014

### AlexPilk

Oh, yes, a truck. And I'm sorry it's actually 1 MN. It's an exercise from some book. I don't know, it seems like a real life situation so this should be solved somehow, because the force pushing it is greater than the friction, otherwise it wouldn't move, so I suppose 0MN-1MN is wrong

4. Oct 31, 2014

### BvU

OK, I'll go for a 2000 ton truck. As long as I don't run into it :)

(Much) more importantly: If the force pushing is greater than the friction, then Newton says it is accelerating.

Constant speed means no acceleration.
No acceleration means no net force.
Very important concept !​

(We can't capitalize: it's considered shouting. But I'd love to do just that!)

The exercise really wants you to do a very simple force balance, and then apply $P = \vec F \cdot \vec v$

and just think of the power you need to accelerate this Leviathan with e.g. 2 m/s2! Or, even worse: how much you must dissipate in an emergency stop at -4 m/s2!
Good thing they have a 50 or 60 ton limit on trucks here...

Last edited: Oct 31, 2014
5. Oct 31, 2014

### AlexPilk

Ok, so the overall force = 0, because it isn't accelerating, and friction = the force pushing it, now I get it. So the force pushing the truck = 1MN, right? And then I just multiply it by the speed?

6. Oct 31, 2014

### BvU

Yep!

7. Oct 31, 2014

### AlexPilk

Thanks a lot! :)

8. Oct 31, 2014

### BvU

So that's 13000 horsepower, just to slug along at 36 km/hour....

9. Oct 31, 2014

### AlexPilk

I'm from Ukraine, our books are written that way. Sometimes you even get negative angles (doesn't make a lot of sense outside geometry) and friction force, because nobody thinks much swapping the values, so yes, a 2000 ton truck :D

10. Oct 31, 2014

### BvU

Sympathize...

11. Oct 31, 2014

### haruspex

I object to the given information that the coefficient of friction is 0.05. Presumably the wheels are in rolling contact with the road, so the coefficient of friction is irrelevant. It should say the coefficient of rolling resistance.

12. Oct 31, 2014

Sustained :)