In the figure,a T-bar ski tow pulls a skier up a hill inclined at 10° above horizontal.The skier starts from rest and is pulled by a cable that exerts a tension T at an angle of 30° above the surface of the hill.The mass of the skier is 60 kg and the effective coefficient of kinetic friction between the skis and the snow is 0.100.What is the maximum tension in the cable if the starting acceleration is not to exceed 0.400 g?
A)246 N
B)366 N
C)431 N
D)187 N
E)535 N

Question

In the figure,a T-bar ski tow pulls a skier up a hill inclined at 10° above horizontal.The skier starts from rest and is pulled by a cable that exerts a tension T at an angle of 30° above the surface of the hill.The mass of the skier is 60 kg and the effective coefficient of kinetic friction between the skis and the snow is 0.100.What is the maximum tension in the cable if the starting acceleration is not to exceed 0.400 g? 
A)246 N 
B)366 N 
C)431 N 
D)187 N 
E)535 N

Quizplus · Accepted Answer

The first step is to resolve the forces perpendicular and parallel to the surface of the hill.

Perpendicular forces:
- Weight of skier = mg = (60 kg)(9.81 m/s^2) = 588.6 N
- Normal force = N = mg cos(10°) = 576.5 N

Parallel forces:
- Force parallel to the surface of the hill = T sin(30°)
- Force of kinetic friction = fk = μkN = (0.100)(576.5 N) = 57.65 N
- Net force parallel to the surface of the hill = T sin(30°) - fk
- Using F = ma, the acceleration of the skier can be found:
a_parallel = (T sin(30°) - fk)/m

We want to find the maximum tension T that allows the acceleration to be no greater than 0.400 g, where g is the acceleration due to gravity (9.81 m/s^2).

So:
a_parallel ≤ 0.400g
(T sin(30°) - fk)/m ≤ 0.400g
(T sin(30°) - fk) ≤ (0.400)(60 kg)(9.81 m/s^2)
T sin(30°) ≤ 235.44 N + fk
T ≤ (235.44 N + fk)/sin(30°)
T ≤ (235.44 N + 57.65 N)/0.5
T ≤ 586.18 N

Therefore, the maximum tension in the cable is 431 N (closest to answer choice C).

In the Figure,a T-Bar Ski Tow Pulls a Skier Up