A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 24.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 60.0 kg, and the coefficient of kinetic friction between the skis and the snow is 0.160. Find the magnitude of the force that the tow bar exerts on the skier.
A skier of mass 83.3 kg comes down a slope of
constant angle 28� with the horizontal.
The acceleration of gravity is 9.8 m/s2 .
What is the force on the skier parallel to
the slope?
Answer in units of N
To find the magnitude of the force that the tow bar exerts on the skier, we need to analyze the forces acting on the skier.
1. Weight: The skier's weight acts vertically downward with a magnitude of Fw = m * g, where m is the mass of the skier (60.0 kg) and g is the acceleration due to gravity (9.8 m/s^2). So, Fw = 60.0 kg * 9.8 m/s^2 = 588.0 N.
2. Normal Force: The normal force acts perpendicular to the surface and balances the weight of the skier. On an inclined plane, the normal force is given by Fn = Fw * cos(θ), where θ is the angle of incline (24.0°). So, Fn = 588.0 N * cos(24.0°) = 532.7 N.
3. Force of kinetic friction: The force of kinetic friction opposes the motion of the skier. It acts parallel to the slope and can be calculated as Fk = μ * Fn, where μ is the coefficient of kinetic friction (0.160) and Fn is the normal force. So, Fk = 0.160 * 532.7 N = 85.1 N.
4. Force from the tow bar: Since the skier is being pulled up the slope at a constant velocity, the force from the tow bar must be equal to the sum of the force of kinetic friction and the component of the weight along the slope (parallel to the slope).
We can find this component of the weight by multiplying the weight by the sine of the angle of incline. So, the component of the weight along the slope is given by Fw_s = Fw * sin(θ) = 588.0 N * sin(24.0°) = 244.2 N.
Therefore, the magnitude of the force that the tow bar exerts on the skier is the sum of the force of kinetic friction and the component of the weight along the slope:
Ftow = Fk + Fw_s
= 85.1 N + 244.2 N
= 329.3 N.
So, the magnitude of the force that the tow bar exerts on the skier is 329.3 N.