Question

A 7.5­kg block is sliding down a wall with constant velocity. The coefficient of static friction between the block and the wall is μ s = 0.70. The coefficient of kinetic friction between the block and the wall is μ k = 0.40. What is the magnitude of the force (in N) pressing the block against the wall? Use g = 9.79 m/s2

Answer 1

Ok, a couple of things have to be accounted for here.  First, since the block is moving relative to the wall we have to use the kinetic coefficient of friction, 0.40.  The second consideration is that since the block is moving at a constant velocity, the acceleration is zero.  This means, by Newton's second Law, that the net force is zero.  So the force of gravity must be equal to the friction force of the wall resisting its fall. This friction force is the product of the normal force (which we are seeking) and the kinetic coefficient of friction. We can then set these two forces equal:

$F_{gravity}=mg \\ \\ F_{friction}=F_{norm} \mu_k \\ \\ mg=F_{norm} \mu_k \\ \\ F_{norm}= \frac{mg}{ \mu_k} \\ \\ F_{norm}= \frac{7.5(9.79)}{0.4}=183.6N$