(a) In the diagram illustrated, a body of mass m slides on an inclined plane. Show that the coefficient Mg of friction between the surfaces in contact is ta...

Answer Details

(a) Showing that \(\mu = \tan\theta\)

The diagram shows a body of mass m on a plane inclined at angle \(\theta\). The forces on it are the weight \(Mg\) acting vertically downward, the normal reaction \(R\) perpendicular to the plane, and the frictional force \(F_p\) acting up the plane (opposing the tendency to slide down).

Resolve the weight into components parallel and perpendicular to the plane:

• Component along the plane (down the slope): \(Mg\sin\theta\)
• Component perpendicular to the plane: \(Mg\cos\theta\)

Perpendicular to the plane there is no motion, so

\[ R = Mg\cos\theta. \]

When the body is just on the point of sliding (or slides down at constant velocity), the frictional force is limiting and balances the component of weight down the plane:

\[ F_p = Mg\sin\theta. \]

But the limiting friction is \(F_p = \mu R\). Therefore

\[ \mu R = Mg\sin\theta. \]

Substituting \(R = Mg\cos\theta\):

\[ \mu\,Mg\cos\theta = Mg\sin\theta \]\[ \mu = \frac{\sin\theta}{\cos\theta} = \tan\theta. \]

Hence the coefficient of friction equals \(\tan\theta\), where \(\theta\) is the angle of repose.

Relative density of the metal

Because the spring obeys Hooke's law, the extension is proportional to the load (force) on it.

Weight of metal in air \(\propto\) extension in air \(= 10.5\,\text{cm}\).

Apparent weight in water \(\propto\) extension in water \(= 6.8\,\text{cm}\).

Upthrust = loss in weight \(\propto (10.5 - 6.8) = 3.7\,\text{cm}\).

The upthrust equals the weight of water displaced, so

\[ \text{Relative density} = \frac{\text{weight in air}}{\text{weight of water displaced}} = \frac{\text{extension in air}}{\text{loss in extension}}. \]\[ \text{R.D.} = \frac{10.5}{10.5 - 6.8} = \frac{10.5}{3.7} = 2.84. \]