# Work done in moving a wheel

The wikipedia article on the wheel says that

If a 100 kg object is dragged for 10 m along a surface with the coefficient of friction μ = 0.5, the normal force is 981 N and the work done (required energy) is (work=force x distance) 981 × 0.5 × 10 = 4905 joules.

Now give the object 4 wheels. The normal force between the 4 wheels and axles is the same (in total) 981 N. Assume, for wood, μ = 0.25, and say the wheel diameter is 1000 mm and axle diameter is 50 mm. So while the object still moves 10 m the sliding frictional surfaces only slide over each other a distance of 0.5 m. The work done is 981 × 0.25 × 0.5 = 123 joules; the work done has reduced to 1/40 of that of dragging.

How did they get the distance of 0.5m? I can't seem to get my head around it.

• The wheel doesn't slide on the floor, but the axle does slide against whatever holds the axle fixed. It's saying that if the cart moves 10m, then the axle slide 0.5m against the cart as the wheels turn. – BowlOfRed Dec 18 '15 at 23:25
• Would that 0.5m be for all the 4 wheels or just 1 wheel? – BlurryPic Dec 18 '15 at 23:30
• Notice that 0.5/10 is 1/20. Also notice the ratio of the wheel diameter to the axle diameter. Finally notice the difference in the coefficients of friction. – dmckee --- ex-moderator kitten Dec 18 '15 at 23:36

The distance the cart moves is $10$m, so how many rotations of the wheels happen over that distance? Well we're told the diameter of the wheels is $1$m so the circumference of the wheels is $\pi$. The number of rotations is therefore:

$$N = \frac{10}{\pi}$$

We're told the diameter of the axle is $0.05$m, so the circumference of the axle is $0.05\pi$m. The distance moved by the axle/bearing contact point is therefore:

$$d = N\times0.05\pi = 10 \times 0.05 = 0.5\,\text{m}$$

And that's where the $0.5$m comes from.

All the wheels have their axle/bearings move $0.5$m but actually the number of wheels doesn't affect the calculation of the work done. Suppose we have $n$ wheels then the load on each wheel bearing is:

$$L = \frac{mg}{n}$$

The frictional force is $L\mu$ and the work is force times distance or $0.5L\mu$. Substituting for $L$ to get the work done per wheel gives:

$$W_\text{wheel} = 0.5\mu\frac{mg}{n}$$

and to get the total work we multiply by the number of wheels $n$ to give:

$$W_\text{total} = 0.5\mu\frac{mg}{n} n = 0.5\mu mg$$

So the number of wheels cancels out and doesn't appear in the equation for the total work.

• Yes you got is it right. This is as precise as it can get. – Vinay5forPrime Dec 19 '15 at 15:00