A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 28.2 ° 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 46.7 kg, and the coefficient of kinetic friction between the skis and the snow is 0.226. Find the magnitude of the force that the tow bar exerts on the skier.
constant velocity= zero acceleration, so force pulling= forcefriction+forcegravitydownslope
force pulling= mg*mu*cosTheta+mgSinTheta
To find 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 (mg): The weight of the skier is given by the equation: weight = mass × gravitational acceleration.
weight = (46.7 kg) × (9.8 m/s^2) = 457.66 N (rounded to two decimal places).
2. Normal force (N): The normal force is the force exerted by a surface perpendicular to the object. On an inclined slope, the normal force can be calculated using the equation: N = weight × cos(θ), where θ is the angle of the slope.
N = 457.66 N × cos(28.2°) = 411.92 N (rounded to two decimal places).
3. Force of Kinetic Friction (Ff): The force of kinetic friction can be calculated using the equation: Ff = coefficient of kinetic friction × N.
Ff = 0.226 × 411.92 N = 93.12 N (rounded to two decimal places).
4. Tension force in the tow bar: The tension force in the tow bar is equal to the sum of the force of kinetic friction and the component force along the slope. Since the skier is being pulled up the slope at a constant velocity, the force along the slope must be equal to the force of kinetic friction.
Tension force = Ff = 93.12 N.
Therefore, the magnitude of the force that the tow bar exerts on the skier is 93.12 N.
To find the magnitude of the force that the tow bar exerts on the skier, we need to consider the forces acting on the skier and find the net force acting on the skier in the direction of motion.
First, let's analyze the forces acting on the skier:
1. The force of gravity (mg): This force acts vertically downward and can be calculated by multiplying the mass of the skier (m) by the acceleration due to gravity (g ≈ 9.8 m/s²).
2. The normal force (N): This force acts perpendicular to the slope and counters the force of gravity. It can be calculated using the component of the weight vector perpendicular to the slope, which is N = mg cos(θ), where θ is the angle of inclination (28.2°).
3. The friction force (f): This force acts parallel to the slope and opposes the motion of the skier. It can be calculated using the coefficient of kinetic friction (μk = 0.226) multiplied by the normal force (f = μkN).
4. The force applied by the tow bar (F): This force is parallel to the slope and helps pull the skier up the slope. This is the force we want to find.
Since the skier is moving up the slope at a constant velocity, the net force acting on the skier in the direction of motion is zero.
The net force in the direction of motion is the vector sum of the force applied by the tow bar (F) and the friction force (f). Therefore:
F - f = 0
Substituting the expression for the friction force:
F - μkN = 0
Now, let's substitute the values given in the problem:
μk = 0.226
m = 46.7 kg
θ = 28.2°
g ≈ 9.8 m/s²
First, we need to calculate the normal force:
N = mg cos(θ)
N = (46.7 kg)(9.8 m/s²) cos(28.2°)
Next, we can substitute the value for N into the equation for the force applied by the tow bar:
F - (0.226)(N) = 0
F - (0.226)[(46.7 kg)(9.8 m/s²) cos(28.2°)] = 0
Now, we can solve for F:
F = (0.226)[(46.7 kg)(9.8 m/s²) cos(28.2°)]
Calculating this expression will give us the magnitude of the force that the tow bar exerts on the skier.