M-10 Power and Friction

APPARATUS: Prony brake, scale, timer, stopwatch, 15 meter tape.

Part A - POWER

 

1.

Measure your horsepower (1 hp = 746 watts) by use of the Prony brake which is mounted on the wall. Try a slow rate that you could keep up all day and another as fast as you can turn the wheel. Figure out what data are required and how to use them. Check with your instructor to be sure you have analyzed the problem correctly. Many calculate the frictional force incorrectly.

2.

Measure the horsepower you develop running up a flight of stairs.

Part B - FRICTION

INTRODUCTION:


 We measure the coefficient of kinetic friction, $\mu$, between rope and wheel of the Prony brake and test whether $\mu$ is independent of velocity and of normal force. (By the way, the Prony brake is named after Gaspard Clair Francois Marie Riche de Prony, who invented it in Paris in 1821 to measure the power of engines. He was the director of the Ecole des Ponts et Chausses, a position he chose over joining Napoleon's army that invaded Egypt.)

Let f = frictional force/unit length and n = normal force/unit length along the rope. For an element $\Delta$s = $r\Delta \theta$ along the rope, let T' and T be the tensions at opposite ends of $\Delta$s. For small $\Delta \theta$, tensions $\vec{T}'$ and $\vec{T}$will be almost antiparallel and differ in magnitude, $T' - T = \Delta T$, only by the frictional force $f\Delta s$. Thus

$$\Delta T = f \Delta s$$ (1)

The reaction of the wheel must supply a normal force $n \Delta s$ sufficient to balance the vector sum $\vec{T}' + \vec{T}$, or $n \Delta s = \vert\vec{T}' + \vec{T}\vert$. But from the figure we see that $\vert\vec{T}' + \vec{T}\vert \simeq T \Delta \theta$ so

$$T \Delta \theta = n \Delta s$$ (2)

Eliminating $\Delta$s between (1) and (2) gives $T \Delta \theta = n \Delta T/f = \Delta T/\mu$since $f = \mu n$. Hence

$$\mu \Delta \theta = \frac{\Delta T}{T} \hspace*{.5in} \mbox{and} \hspace*{.5in} \mu d \theta = \frac{dT}{T}~~.$$

For the Prony brake, we integrate between and $\pi$:

$$\mu \int_0^{\pi} d \theta = \int_{T_1}^{T_2} dT \over T \hsp... ...2in} \mu = \frac{1}{\pi} \ell n \left( \frac{T_2}{T_1} \right)$$

Note #1: While for the Prony brake, $\theta = \pi$, the general case, $\theta \neq \pi$, corresponds to the ``wrapped string tension'' used e.g. for ``warping a ship''. Then $dT/T = \mu d \theta$ integrates to

$$\ell n (T/T_0) = \mu \theta \hspace*{.2in} \mbox{and hence} \hspace*{.2in} T = T_0 e^{\mu \theta}$$

which holds also for a slipping string if $\mu = \mu_k$.

SUGGESTIONS FOR PART B:

 

1.

Calculate your value of $\mu$.

2.

Test whether $\mu$ is independent of velocity by turning at different speeds and noting that neither T₂ nor T₁ changes appreciably.

3.

Test whether $\mu$ is independent of the normal force by using a different tension setting for the brake. (Use different link setting in the chain to vary the tension).

QUESTIONS:

 

1.

For a stationary wheel the two spring balances will not read the same if

(a)

there are zero errors in the balances,

(b)

there is friction in the wheel bearing.

(c)

the handle for turning the brake is not directly above or below the wheel's axis.

Discuss the errors resulting from these effects.

2.

How might one understand a change in $\mu$ as the normal force increases in #3 of Part B?

Note #2:


  Friction arises from intermolecular forces between the atoms (or molecules) of the two surfaces, and so frictional forces are very complex. It is therefore surprising that the simple empirical relations for friction (e.g. that $\mu$ is a constant) hold as well as they do.