A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 25.9° 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 55.3 kg, and the coefficient of kinetic friction between the skis and the snow is 0.141. Calculate the magnitude of the force that the tow bar exerts on the skier.
To calculate the magnitude of the force that the tow bar exerts on the skier, we need to consider the forces acting on the skier.
1. Weight: The weight of the skier can be calculated using the formula W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity. Assuming g = 9.8 m/s², we have:
W = 55.3 kg * 9.8 m/s² = 541.94 N
2. Normal force: The normal force acts perpendicular to the slope and counterbalances the weight. On an inclined plane, the normal force can be calculated using the formula N = W * cos(θ), where θ is the angle of inclination. Therefore, we have:
N = 541.94 N * cos(25.9°) = 485.49 N
3. Friction force: The friction force is parallel to the slope and can be calculated using the formula F_friction = μ * N, where μ is the coefficient of kinetic friction. Thus, we have:
F_friction = 0.141 * 485.49 N = 68.5 N
4. Applied force: The applied force by the tow bar is parallel to the slope and opposes the friction force to maintain a constant velocity. Therefore, the magnitude of the force applied by the tow bar can be calculated by summing up the forces acting on the skier.
F_applied = F_friction = 68.5 N
Thus, the magnitude of the force that the tow bar exerts on the skier is 68.5 Newtons.
To calculate the magnitude of the force that the tow bar exerts on the skier, we need to consider the forces acting on the skier.
Let's first break down the forces acting on the skier:
1. The force applied by the tow bar (parallel to the slope): F_applied
2. The force of gravity acting vertically downward: F_gravity = m * g, where m is the mass of the skier (55.3 kg) and g is the acceleration due to gravity (9.8 m/s^2).
3. The normal force perpendicular to the slope: F_normal
4. The force of kinetic friction opposing the skier's motion: F_friction = coefficient of kinetic friction * F_normal, where the coefficient of kinetic friction is given as 0.141.
Since the skier is being pulled up the slope with a constant velocity, the net force acting on the skier is zero, meaning the sum of all forces must be zero.
In the vertical direction, we have:
F_normal - F_gravity * cos(25.9°) = 0
Simplifying, we find:
F_normal = F_gravity * cos(25.9°)
In the horizontal direction, we have:
F_applied - F_friction - F_gravity * sin(25.9°) = 0
Now, let's substitute the known values:
F_applied - (0.141 * F_normal) - (55.3 kg * 9.8 m/s^2 * sin(25.9°)) = 0
Substituting the value of F_normal from the first equation, we get:
F_applied - (0.141 * F_gravity * cos(25.9°)) - (55.3 kg * 9.8 m/s^2 * sin(25.9°)) = 0
Now, we can solve for F_applied:
F_applied = (0.141 * F_gravity * cos(25.9°)) + (55.3 kg * 9.8 m/s^2 * sin(25.9°))
Substituting the values:
F_applied ≈ (0.141 * (55.3 kg * 9.8 m/s^2) * cos(25.9°)) + (55.3 kg * 9.8 m/s^2 * sin(25.9°))
Calculating the value, we find:
F_applied ≈ 504.4 N (rounded to one decimal place)
Therefore, the magnitude of the force that the tow bar exerts on the skier is approximately 504.4 N.