I'm a pure mathematician by trade, but have been teaching myself classical mechanics. I've got to the chapter on Work, Energy and Power and I've found an example that is causing me some problems.


A van of mass 1250 kg is travelling along a horizontal road. The van's engine is working at 24 kW. The constant force of resistance to motion has magnitude 600 N. Calculate:

  1. The acceleration of the van when it is travelling at 6 m/s.
  2. The maximum speed of the van.


Part 1 is fine. We know that $P = Fv$ and so $24000 = 6F$. It follows that $F = 4000$ N. To find the acceleration, we use $F=ma.$ The total resulting force is the traction minus the resistance, in other words $F = 4000 - 600 = 3400$ N. The mass is 1250 kg and so $3400 = 1250a$. It now follows that $a=2.72$ m/s/s.

Part 2 is the part that causes me problems. The maths isn't the problem, it's the way it's used. At maximum speed the acceleration is zero and so the resultant horizontal force will be zero. The book says that $T' = 600$. There are two problems here: I'm used to a prime denoting differentiation. I guess it just means the new tractive force. But the resistance force is working against the direction of motion, so shouldn't we have $T' = -600$?

Running with $T' = 600$, the book then uses $P = Fv$ to get $24000=600v$ and so $v=40$ m/s. I can see that using $T'=-600$ would give a negative velocity, which is clearly untrue.

In short

I don't see why the fact that at maximum speed there will be no acceleration, so the resultant horizontal force will be zero leads to us using $T'=600$ when $T > 0$ was the forwards tractive force and $-600 < 0$ was the resistance force.

  • $\begingroup$ Hi Fly by Night. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Aug 30 '16 at 13:19

I don't think you initially defined $T$, so I'm not entirely sure what it originally used for. The overall idea, however, is that because the newt force is zero (and the resistive force stays the same, i.e. $-600$, then the new forward force has to be $T' = +600$.

Physics is often (e.g. besides high-energy) much more pragmatic than mathematics per se -- one expression of this is the mixed use of things like primes, or superscripts, etc. Similarly, negative signs often aren't tracked quite as closely (as they should be). Because the positive or negative are purely conventional, and it is conceptually clear that the resulting force (and velocity) need to be opposite the resistive force -- then you can just give them the appropriate sign. It is, of course, better to treat them accurately and consistently, but note that this is a common location of hand-waving in introductory level textbooks.

  • $\begingroup$ Sorry: $T$ is used for the tractive force, i.e. the total pulling power of the vehicle. $\endgroup$ – Fly by Night Dec 26 '12 at 22:59
  • $\begingroup$ I have noticed quite a lot of notation misuses in the textbook I've been reading. For example they use $F$ for a force and $F$ for the magnitude of that force. I would use ${\bf F}$ and $||{\bf F}||$ respectively. I think I have resolved the problem. As you say, the net residence is $-600$ N and so the engine must produce a net force of $+600$ N in order to maintain constant speed. That is the force used in the calculations. I can understand why students struggle so much: some of the textbooks leave a lot unsaid. Thanks a lot for taking the time to reply, it's much appreciated $\endgroup$ – Fly by Night Dec 26 '12 at 23:00
  • $\begingroup$ Happy to help!! $\endgroup$ – DilithiumMatrix Dec 27 '12 at 21:44

Three points.

Firstly, remember that power and work are not vector quantities.

You said "I can see that using $T′=−600T′=−600$ would give a negative velocity, which is clearly untrue."

But since we are 'playing this game' in a straight line, and vector dot products are not really needed, then only positive values of force and velocity are needed, to simplify the maths. It would not be a valid statement to use a negative value of force in the 'scalar equation' of $P= Fv$.

Secondly, if T is the symbol for the Tractive force of the motor, then the books use of T' is a bit confusing. I would guess that T' actually stands for the Resistive force acting against the Tractive force. It would have been better for them to have used a completely different symbol, such as R for resistance say.

So in Part 2 of the answer, where the Total Force acting on the van is zero, it is true to say that the VALUES of the Traction force and the resistive force are both 600. i.e. $T = 600~\mathrm{N}$ and also $T' = 600~\mathrm{N}$.

And thirdly, because of the two similar symbols being used for two different quantities, you have accidently referred to T', rather than T in the last part of your question . . .

The $F$ in the formula of $P = Fv$ should really be changed to $P = T*v$ since we are using the symbol T for the tractive force that is generating the output power of the motor.

So "using $T′=−600T′=−600$ would give a negative velocity" is not a valid statement because in that equation T' is not the quantity to be used. It is $T$ that is being used, the tractive force, not the resistive force.


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