A skier of mass 59.5 kg comes down a slope of constant angle 1◦ with the horizontal.
The acceleration of gravity is 9.8 m/s^2
i. What is the force on the skier parallel to the slope?
Answer in units of N.
ii. What force normal to the slope is exerted by the skis?
Answer in units of N.
i. Well, if the skier is coming down a slope, gravity is definitely pulling him in that direction. So the force parallel to the slope is just the force of gravity, which can be calculated by multiplying the mass of the skier (59.5 kg) by the acceleration due to gravity (9.8 m/s^2). So, the force on the skier parallel to the slope is approximately 582.1 N.
ii. The force normal to the slope is basically the force exerted by the skis in the perpendicular direction. It is equal in magnitude but opposite in direction to the force of gravity acting on the skier. So, it is also around 582.1 N, but with opposite direction.
To solve this problem, we need to consider the forces acting on the skier.
i. Force parallel to the slope (Fpar) can be calculated using the formula:
Fpar = m * g * sin(θ)
where:
m = mass of the skier = 59.5 kg
g = acceleration due to gravity = 9.8 m/s^2
θ = angle of the slope = 1° (converted to radians, 1° * π / 180°)
Substituting the values into the formula:
Fpar = 59.5 kg * 9.8 m/s^2 * sin(1° * π / 180°)
≈ 9.10 N (rounded to two decimal places)
Therefore, the force on the skier parallel to the slope is approximately 9.10 N.
ii. The force normal to the slope (Fnorm) can be calculated using the formula:
Fnorm = m * g * cos(θ)
where:
m = mass of the skier = 59.5 kg
g = acceleration due to gravity = 9.8 m/s^2
θ = angle of the slope = 1° (converted to radians, 1° * π / 180°)
Substituting the values into the formula:
Fnorm = 59.5 kg * 9.8 m/s^2 * cos(1° * π / 180°)
≈ 588.42 N (rounded to two decimal places)
Therefore, the force normal to the slope exerted by the skis is approximately 588.42 N.
To find the force on the skier parallel to the slope, we need to calculate the component of the weight of the skier that acts parallel to the slope.
i. The force on the skier parallel to the slope can be found using the formula: F_parallel = m * g * sinθ, where m is the mass of the skier, g is the acceleration due to gravity, and θ is the angle of the slope.
Given:
Mass of the skier (m) = 59.5 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Angle of the slope (θ) = 1 degree
First, we need to convert the angle from degrees to radians:
θ_radians = θ * (π / 180)
θ_radians = 1 * (π / 180)
Now we can plug in the values to calculate the force parallel to the slope:
F_parallel = 59.5 kg * 9.8 m/s^2 * sin(1 * (π / 180))
Using a calculator, you can find the value of sin(1 * (π / 180)) ≈ 0.01745
F_parallel ≈ 59.5 kg * 9.8 m/s^2 * 0.01745
Now, you can calculate the force parallel to the slope.
ii. The force normal to the slope is the component of the weight of the skier that acts perpendicular to the slope.
The force normal to the slope can be found using the formula: F_normal = m * g * cosθ
Using the same values as given in part i:
F_normal = 59.5 kg * 9.8 m/s^2 * cos(1 * (π / 180))
Again, using a calculator, you can find the value of cos(1 * (π / 180)) ≈ 0.99985
F_normal ≈ 59.5 kg * 9.8 m/s^2 * 0.99985
Now, you can calculate the force normal to the slope.
1.Fs = mg*sin A = 583.1*sin1o=10.18 N.
2. Fn = mg*cos1o = 583.1*cos1o=583.0 N.