A 25-kg uniform ladder 8.0 m long is leaning against a frictionless wall at an angle of 75 degrees above the horizontal.A 40-kg boy climbs 6.0 m up the ladder.What is the magnitude of the frictional force exerted on the ladder by the floor?
A)200 N
B)110 N
C)280 N
D)140 N
E)550 N

Question

Quizplus · Accepted Answer

First, we need to find the force exerted by the ladder on the wall, which can be found using the equation F_h = mg / sinθ, where F_h is the horizontal force exerted by the ladder, m is the mass of the ladder, g is the acceleration due to gravity, and θ is the angle the ladder makes with the horizontal. 
Plugging in the given values, we get F_h = (25 kg) x (9.81 m/s^2) / sin(75°) = 235.5 N

Next, we need to find the force exerted by the ladder on the floor. This can be found using Newton's third law, which states that for every action, there is an equal and opposite reaction. Therefore, the force exerted by the ladder on the floor is equal in magnitude to the force exerted by the floor on the ladder.

To find the force exerted by the boy on the ladder, we can use the equation F_v = mg, where F_v is the vertical force exerted by the boy. Plugging in the given values, we get F_v = (40 kg) x (9.81 m/s^2) = 392.4 N.

Next, we need to find the horizontal and vertical components of the force exerted by the boy on the ladder. The horizontal component can be found using the equation F_h = F_v x tanθ, where θ is the angle the ladder makes with the horizontal. Plugging in the given values, we get F_h = (392.4 N) x tan(75°) = 104.5 N.

Now that we have the horizontal components of both the ladder and the boy, we can use Newton's first law to find the frictional force exerted by the floor, which is equal in magnitude to the horizontal component of the net force on the ladder. 
Net horizontal force = F_h (ladder) - F_h (boy) = 235.5 N - 104.5 N = 131 N. Therefore, the frictional force exerted by the floor is 131 N, which is closest to option B (110 N).

A 25-Kg Uniform Ladder 8