# Help With Firefighter Test Question

1. Sep 12, 2006

### spkspkspkspk

Ok, maybe this is simple and I'm overthinking it...

The question asked "Which cart would be easier to push"

Four carts were shown. They are basic flat carts on four wheels.
Cart 1: All four wheels small (4" diameter)
Cart 2: Front wheels small (4" D), back wheels large (12" D)
Cart 3: Front wheels large (12" D), back wheels small (4" D)
Cart 4: All four wheels large (12" D)

This isn't a question asking any specifics - just asks which would be easier to push. I don't want to overthink the weight of the wheels, and the difference between getting the load started versus maintaining momentum.

My thought was that the larger wheels front and back would be "easier to push" since a larger radius would require less effort. Am I nuts? Been out of college for 10 years now and can't think of the right formula or principle that applies.

2. Sep 12, 2006

### physicsgal

i'd go with all 4 large wheels myself.

~Amy

3. Sep 12, 2006

### Andrew Mason

There are a few assumptions that are being made here. In asking which is easier to push, I think you have to assume that they all have the same total mass. It does not tell you how the wheels are designed, so we have to assume that the big wheels have the same mass as the smaller wheels.

It is not clear what the question is asking. "Easier to push" could mean: "easier to get moving" (ie. easier to overcome friction") or "uses less force to achieve the same acceleration" which is equivalent to "uses less energy to achieve the same speed". I am assuming that it is asking the latter: what configuration results in the greatest change in translational speed for a given force applied over a given distance.

Assuming there is no work done in overcoming friction, there should be no difference between the force required to accelerate the various configurations if they all have the same total mass.

But let's look at it from an energy perspective: If you apply a force F for a distance d the kinetic energy acquired by the cart is KE = W = Fd. This is in the form of rotational energy of the wheels and translational energy of the cart centre of mass. For each of the wheels the rotational energy is $\frac{1}{2}I\omega^2$ where $I = \frac{1}{2}mR^2$. Note also that $\omega = v/R$ where v = the translational speed of the cart centre of mass. So the total energy is (m_c = total cart mass; m_w = mass of each wheel):

$$KE = \frac{1}{2}m_cv^2 + 4\frac{1}{4}m_wR^2v^2/R^2 = \frac{1}{2}m_cv^2 + m_wv^2$$

If they all have the same total cart mass and each wheel has the same mass, the energy needed to reach a certain speed is independent of the wheel size.

AM

4. Sep 12, 2006

### spkspkspkspk

Awesome...

Thank you. This is excellent.

I am faced with the problem that only one answer can be right, and 'all of the above' is not an answer. Sigh.

What if we considered the other intention - that it is asking "which is easier to overcome friction"?

Any thoughts? Again, assuming all wheel options have the same mass.

Maybe this is really about rolling resistance?

Last edited: Sep 12, 2006
5. Mar 14, 2008

### bosak

Wheel size

Did we ever get an answer on the cart/wheel question. I have a similar quetion with a heavy steel gate on metal rollers on steel tracks. Would a 4" wheel or a 12" wheel provide an individual with an easier time opening and closing?

Last edited by a moderator: Mar 14, 2008