A ladder (L = 7.80 m) of weight WL = 360 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 889 N, stands 6.40 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 = 54.0°.)
blackpeoplemeet. com this should help
To find the minimum value for the coefficient of static friction between the ladder and the ground, we need to consider the forces acting on the ladder and the conditions for equilibrium.
First, let's draw a free-body diagram of the ladder. We have the weight of the ladder acting vertically downward at its center with a magnitude of 360 N. The normal force exerted by the wall acts at right angles to the wall and balances the vertical component of the ladder's weight. There is also a horizontal component of the ladder's weight acting due to the angle θ.
The horizontal force acting on the ladder is the frictional force between the ladder and the ground, which resists the tendency of the ladder to slide. This force depends on the coefficient of static friction (μ) between the ladder and the ground.
Now, let's consider the conditions for equilibrium. Since the ladder is not slipping, the net force and the net torque acting on it must be zero.
1. Net vertical force:
The vertical component of the ladder's weight is balanced by the normal force exerted by the wall:
N = Wladder_vertical
N = Mg
Mg = WLadder_vertical
Mg = WLadder * cos(θ)
2. Net horizontal force:
The horizontal component of the ladder's weight is balanced by the frictional force:
f = WLadder_horizontal
f = Mg * sin(θ)
3. Net torque:
The torque due to the ladder's weight is balanced by the torque due to the firefighter's weight:
τ = Wfirefighter * d
τ = Mg * d
τ = Wladder * L/2 * sin(θ)
τ = Mg * L/2 * sin(θ)
Since the ladder is in equilibrium, the net torque is zero:
τ = 0
Mg * L/2 * sin(θ) = 0
Mg * L/2 * sin(θ) = Wfirefighter * d
Now, substitute the values given in the problem:
M = 360 N
L = 7.80 m
Wfirefighter = 889 N
d = 6.40 m
θ = 54.0°
Plug in these values to the equations and solve for μ:
Mg = WLadder * cos(θ)
360 N * g = 360 N * cos(54.0°)
g = 9.8 m/s^2 (acceleration due to gravity)
f = Mg * sin(θ)
f = 360 N * g * sin(54.0°)
τ = Mg * L/2 * sin(θ)
Mg * L/2 * sin(θ) = Wfirefighter * d
360 N * 7.80 m * sin(54.0°) = 889 N * 6.40 m
Now you can calculate the minimum value for the coefficient of static friction (μ) by dividing the frictional force (f) by the normal force (N):
μ = f / N