What is the smallest of the force F (indicated with the arrow perpendicular to the wall on the diagram below) that the 2.0 kg block will not slide down the wall? The coefficient of static friction between the block and wall is 0.2.
Let's start by drawing a free-body diagram for the block:
The gravitational force acting on the block is given by:
$F_g = mg$
where $m$ is the mass of the block and $g$ is the acceleration due to gravity. Substituting in the given values, we get:
$F_g = (2.0 kg)(9.81 m/s^2) = 19.62 N$
The force of static friction opposing the gravitational force is given by:
$F_f = \mu_s F_N$
where $\mu_s$ is the coefficient of static friction and $F_N$ is the normal force. The normal force is equal to the gravitational force in this case, since the block is resting on a vertical wall. Substituting in the given values, we get:
$F_f = (0.2)(19.62 N) = 3.924 N$
We can now use the fact that the force $F$ must balance the gravitational force and the force of static friction in order to keep the block from sliding down. That is:
$F \sin \theta = F_f$
where $\theta$ is the angle between the wall and the vertical. Since the block is resting on a vertical wall, we have $\theta = 90^\circ$. Substituting in the values we found for $F_f$, we get:
$F \sin 90^\circ = 3.924 N$
which simplifies to:
$F = 3.924 N$
Thus, the smallest force $F$ that will keep the block from sliding down the wall is 3.924 N.