Question

An iron block, of mass 10 Kg, rests on a wooden plane inclined at 30 0 to the horizontal. It is found that the least force parallel to the plane which causes the block to slide up the plane is 10 N. Calculate the coefficient of sliding friction between wood and iron.

Answer 1

Let F be the modulus of the least force parallel to the plane, $\alpha$ be the angle of inclination.

Let the x-axis be directed along the plane upwards, y-axis be perpendicular to x-axis and also directed upwards. If F is the least force needed to begin the motion, the acceleration will be 0.

x-axis: $F -mg\sin\alpha -\mu N = 0.$

y-axis: $N - mg\cos\alpha = 0,$ therefore $N = mg\cos\alpha.$

So $F - mg\sin\alpha -\mu mg\cos\alpha = 0 \Rightarrow \mu = \dfrac{F-mg\sin\alpha}{mg\cos\alpha} = \dfrac{10\,\mathrm{N} - 10\,\mathrm{kg}\cdot10\,\mathrm{N/kg}\cdot\sin30^\circ}{10\,\mathrm{kg}\cdot10\,\mathrm{N/kg}\cdot\cos30^\circ} < 0 !!!$

So it is a mistake in the problem situation, because 10 N is less than the component of gravitational force.

If F = 100 N, then

$\mu = \dfrac{F-mg\sin\alpha}{mg\cos\alpha} = \dfrac{100\,\mathrm{N} - 10\,\mathrm{kg}\cdot10\,\mathrm{N/kg}\cdot\sin30^\circ}{10\,\mathrm{kg}\cdot10\,\mathrm{N/kg}\cdot\cos30^\circ} = 0.58$ , it's a realistic value.