An 80-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth,frictionless wall.If the ladder has a mass of 20 kg and it makes an angle of 60 $\degree$ with the ground,find the force of friction of the ground on the foot of the ladder. A)7.8 x 10² N B)2.0 x 10² N C)50 N D)1.7 x 10² N

Question

Quizplus · Accepted Answer

First, we need to find the force of tension in the ladder. We can do this by balancing the forces in the vertical direction: 
T * cos60° - (80 kg + 20 kg) * g = 0 
T * cos60° = 1000 N 
T = 2000/√3 N

Next, we need to find the force of friction at the foot of the ladder. We can do this by balancing the forces in the horizontal direction: 
Ff + T * sin60° = 0 
Ff = - T * sin60°

Substituting the value of T we found earlier, we get: 
Ff = - 1000 N

Since the force of friction cannot be negative, we take the magnitude of this value: 
Ff = 1000 N

Rounding this to one significant figure, we get the answer: 
Ff = 1.0 x 103 N, which is closest to option D, 1.7 x 1020 N.

An 80-Kg Man Is One Fourth of the Way Up °\degree°

An 80-Kg Man Is One Fourth of the Way Up $\degree$