A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 23.0° 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 55.3 kg, and the coefficient of kinetic friction between the skis and the snow is 0.110. Calculate the magnitude of the force that the tow bar exerts on the skier.

Please start with a free-body diagram, then resolve forces parallel and normal to the slope.

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Assuming the skier (the round object) is moving up the slope, then the friction force is towards downslope.

The towbar force is therefore, equating forces along the slope,
F=mg sin(θ)+μ(mg cos(θ))
substitute
m=55.3 kg
g=9.81 m/s^2
θ=23°
μ=0.110

I get a little less than 270 N.

Why did the skier bring a tow bar to the slopes? Because he wanted to pull some sick moves! Anyway, let's solve this problem.

First, we need to find the normal force acting on the skier. Since the skier is on an inclined plane, the normal force is not equal to the skier's weight. It's actually equal to the component of the skier's weight that is perpendicular to the slope.

The equation for the normal force is: N = mg cos(θ), where m is the mass of the skier (55.3 kg), g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the slope (23.0°).

N = (55.3 kg)(9.8 m/s^2) cos(23.0°)
N = 525.1 N

Next, we can find the force of friction acting on the skier. The equation for the force of friction is: f = μN, where μ is the coefficient of kinetic friction between the skis and the snow (0.110).

f = (0.110)(525.1 N)
f = 57.8 N

Lastly, we can find the magnitude of the force that the tow bar exerts on the skier. Since the skier is moving at a constant velocity, the force exerted by the tow bar must be equal in magnitude and opposite in direction to the force of friction.

F = -f
F = -57.8 N

So, the magnitude of the force that the tow bar exerts on the skier is 57.8 N (opposite in direction to the force of friction). Keep skiing and stay funny, my friend!

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: The weight of the skier can be calculated as the product of the mass and acceleration due to gravity. The formula is given by:
Weight = mass * gravitational acceleration
Weight = 55.3 kg * 9.8 m/s^2
Weight = 541.94 N

2. Normal force: The normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it. On an inclined slope, the normal force is equal in magnitude and opposite in direction to the component of the weight perpendicular to the slope. The formula for the normal force is given by:
Normal force = Weight * cos(θ)
where θ is the angle of inclination.
Normal force = 541.94 N * cos(23°)
Normal force = 486.71 N

3. Friction force: The friction force opposes the motion of the skier. The formula for the friction force is given by:
Friction force = coefficient of friction * Normal force
Friction force = 0.110 * 486.71 N
Friction force = 53.54 N

Since the skier is being pulled up the slope at a constant velocity, the net force acting on the skier is zero. Therefore, the force applied by the tow bar must be equal in magnitude and opposite in direction to the sum of the weight component parallel to the slope and the friction force.

Force by tow bar = Weight * sin(θ) + Friction force
Force by tow bar = 541.94 N * sin(23°) + 53.54 N
Force by tow bar = 229.23 N + 53.54 N
Force by tow bar ≈ 282.77 N

Therefore, the magnitude of the force that the tow bar exerts on the skier is approximately 282.77 N.

To calculate the magnitude of the force that the tow bar exerts on the skier, we need to break down the forces acting on the skier and use the equilibrium condition.

1. Draw a free body diagram: Draw a diagram with the skier and all the forces acting on them. In this case, the forces acting on the skier are the force of gravity (mg), the normal force (N), the force of friction (f), and the force applied by the tow bar (F_bar).

2. Resolve the force of gravity: The force of gravity can be resolved into two components based on the inclined plane. It can be split into a component parallel to the slope (mg*sinθ) and a component perpendicular to the slope (mg*cosθ), where θ is the angle of inclination.

3. Calculate the normal force: The normal force is the force exerted by the surface on the skier, perpendicular to the slope. It is equal in magnitude and opposite in direction to the perpendicular component of the force of gravity (N = mg*cosθ).

4. Calculate the force of friction: The force of friction can be determined using the equation f = μ*N, where μ is the coefficient of kinetic friction and N is the normal force.

5. Apply the equilibrium condition: Since the skier is moving at a constant velocity, the net force acting on them is zero. This means that the force applied by the tow bar must equal the force of friction to maintain equilibrium.

6. Calculate the force applied by the tow bar: Set the force of friction equal to the force applied by the tow bar (F_bar = f) and substitute the values into the equation.

Using the given information:
- Mass of the skier (m) = 55.3 kg
- Angle of inclination (θ) = 23.0°
- Coefficient of kinetic friction (μ) = 0.110

Now we can plug these values into the equations and solve for the force applied by the tow bar:

1. Resolve the force of gravity: mg*sinθ = (55.3 kg)(9.8 m/s^2)(sin 23.0°) = 237.75 N
2. Calculate the normal force: N = (55.3 kg)(9.8 m/s^2)(cos 23.0°) = 501.86 N
3. Calculate the force of friction: f = (0.110)(501.86 N) = 55.20 N
4. Apply the equilibrium condition: F_bar = f = 55.20 N

Therefore, the magnitude of the force that the tow bar exerts on the skier is 55.20 N.