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

We can start by drawing a free body diagram of the skier.

There are a few forces acting on the skier: the tension force from the cable, the weight force, and the force of kinetic friction opposing the skier's motion.

We can resolve the tension force into its horizontal and vertical components, using the angle given in the problem.

Next, we can use Newton's second law, ΣF = ma, to write an expression for the acceleration of the skier.

ΣF = ma
Tcos(30°) - mg*sin(10°) - μk*mg*cos(10°) = ma

We want to find the maximum tension, so we can rearrange this equation to solve for T:

Tcos(30°) = ma + mg*sin(10°) + μk*mg*cos(10°)
T = (ma + mg*sin(10°) + μk*mg*cos(10°))/cos(30°)

We're given that the maximum acceleration is 0.400 g, so we can substitute that in for a:

T = (0.4mg + mg*sin(10°) + 0.1mg*cos(10°))/cos(30°)

Plugging in numbers gives us:

T = (0.4*60*9.81 + 60*9.81*sin(10°) + 0.1*60*9.81*cos(10°))/cos(30°)
T = 431 N

So the maximum tension in the cable is 431 N (choice C).

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