Question

A block of wood of mass 150 g rests on an inclined plane. If the coefficient of static friction between the surfaces in contact is 0.30, find (a) the greatest angle to which the plane may be tilted without the block slipping, (b) the force parallel to the plane necessary to prevent slipping when the angle of the plane with the horizontal is 30 degrees

Answer 1

(a) Let us determine the projections of forces on the axis directed along the plane (x) and on the axis perpendicular to plane (y). If the angle between the plane and the horizontal is $\alpha$ and the block doesn't slip, then

$y: mg\cos\alpha-N = 0 \;\Rightarrow \; N = mg\cos\alpha, \\ x: mg\sin\alpha - F_{\text{fr}} = 0.$

For the greatest angle the friction force will be $F_{fr} = \mu N = \mu mg\cos\alpha.$ Therefore, $mg\sin\alpha = \mu mg\cos\alpha\; \Rightarrow \; \tan\alpha = \mu, \; \alpha = \arctan\mu = \arctan 0.30 \approx 16.7^\circ.$

(b) Let us determine the projections of forces in this case.

$y: mg\cos\beta-N = 0 \;\Rightarrow \; N = mg\cos\beta, \\ x: mg\sin\beta - F_{\text{fr}} - F = 0.$

In this case $F_{fr} = \mu N = \mu m g\cos\beta,$ so $F = mg\sin\beta - \mu m g \cos\alpha = mg(\sin\beta - \mu\cos\alpha) = 0.150\,\mathrm{kg}\cdot9.81\,\mathrm{N/kg}\cdot(\sin30^\circ- 0.30\cos30^\circ) = 0.35\,\mathrm{N}.$