A 75.0 kg ladder that is 3.00 m long is placed against a wall at an angle theta. The center of gravity of the ladder is at a point 1.20 m from the base of the ladder. The coefficient of static friction at the base of the ladder is 0.800. There is no friction between the wall and the ladder. What is the vertical force of the ground on the ladder?
A) 625 N
B) 640 N
C) 735 N
D) 832 N
E) 900 N

Question

A 75.0 kg ladder that is 3.00 m long is placed against a wall at an angle theta. The center of gravity of the ladder is at a point 1.20 m from the base of the ladder. The coefficient of static friction at the base of the ladder is 0.800. There is no friction between the wall and the ladder. What is the vertical force of the ground on the ladder? 
A) 625 N 
B) 640 N 
C) 735 N 
D) 832 N 
E) 900 N

Quizplus · Accepted Answer

First, we need to find the force of gravity acting on the ladder. Using the formula Fg = mg, where m is the mass of the ladder and g is the acceleration due to gravity, we get:

Fg = (75.0 kg)(9.81 m/s^2) = 736 N

Next, we need to find the force acting to the left on the ladder due to friction. This force will be equal to the force of friction, which we can find using the formula Ff = μN, where μ is the coefficient of static friction and N is the normal force acting on the ladder. The normal force can be found using the fact that the ladder is in static equilibrium, so the sum of the forces in the y-direction must be zero. Thus:

N + Fgy = Ffy
N = Ffy - Fgy

where Fgy is the force of gravity acting in the y-direction and Ffy is the force of the ground on the ladder in the y-direction. Since the ladder is not moving in the y-direction, Ffy = N. The force of gravity in the y-direction can be found using trigonometry:

Fgy = (mg/L) * x = [(75.0 kg)(9.81 m/s^2)/3.00 m] * 1.20 m = 294.3 N

where L is the length of the ladder and x is the distance from the base of the ladder to the center of gravity. Plugging in the numbers, we get:

N = Ffy = Fgy - μN
N = 294.3 N - 0.800N
N = 612.5 N

Finally, we can find the force of the ground on the ladder in the x-direction using the fact that the ladder is in static equilibrium, so the sum of the forces in the x-direction must be zero. Thus:

Fx = Ff = μN = 0.800(612.5 N) = 490 N

The vertical force of the ground on the ladder is then given by the Pythagorean theorem:

F = sqrt(Fx^2 + Ffy^2) = sqrt((490 N)^2 + (612.5 N)^2) = 735 N

Therefore, the answer is (C).

A 750 Kg Ladder That Is 3