There are only two horizontal forces in this problem, at the wall and at the ground.
Therefore they are equal and opposite, call the magnitude F
now balance vertical forces
886 + 350 = force down = 1236 N
that is therefore the normal force up
at the ladder base
the max friction force is thus
1236 mu
now take moments about top of ladder
clockwise:
1236 * 7.5 cos 51
counterclockwise
1236 mu *7.5 sin 51 + 350 *3.75 cos 51 + 886 *1*cos 51
set those equal and solve for mu
by the way look at the old questions answered below
A ladder (ℓL = 7.50 m) of weight WL = 350 N leans against a smooth vertical wall. The term "smooth" means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is 886 N, stands 6.50 m up from the bottom of the ladder (this distance goes along the ladder, it is not the vertical height). Assume that the ladder's weight acts at the ladder's center, and neglect the hose's weight. What is the minimum value for the coefficient of static friction between the ladder and the ground, so that the ladder (with the fireman on it) does not slip? (Assume θ = 51.0°.)
1 answer
1 answer