A boy pulls a sled of mass 5.0 kg with a rope that makes an 60.0° angle with respect to the
horizontal surface of a frozen pond. The boy pulls on the rope with a force of 10.0 N; and the
sled moves with constant velocity. What is the coefficient of friction between the sled and the
ice?
Constant velocity means that the forces add up to zero.
Do a force balance in the direction of motion. It says:
10.0 cos 60 = friction force = (M g - 10.0 sin 60)*Uk
Solve for Uk, the coefficient of (kinetic) friction.
The reason for the -10.0 sin 60 term is that the upward pull on the rope reduces the normal force of the sled onto the ice. That term gets subtracted from the sled's weight.
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To find the coefficient of friction between the sled and the ice, we need to first analyze the forces acting on the sled.
1. The gravitational force (weight) acts vertically downward with a magnitude given by:
Fg = m * g,
where m is the mass of the sled and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The force applied by the boy pulls the sled forward. Let's call this force F.
3. The horizontal component of the gravitational force and the friction force act in the opposite direction and cancel each other out, resulting in constant velocity. The horizontal component of the gravitational force is given by:
Fgx = Fg * sinθ,
where θ is the angle between the rope and the horizontal surface.
4. The frictional force is given by:
Ff = μ * Fn,
where μ is the coefficient of friction and Fn is the normal force (the force exerted by the surface perpendicular to it).
Since the sled moves with constant velocity, the force applied by the boy (F) must be equal in magnitude and opposite in direction to the frictional force (Ff).
Now, let's break down the forces in the vertical and horizontal directions:
1. Vertical forces:
The normal force (Fn) is equal in magnitude and opposite in direction to the vertical component of the gravitational force:
Fn = Fgy = Fg * cosθ.
2. Horizontal forces:
The force applied by the boy (F) is equal in magnitude and opposite in direction to the horizontal component of the gravitational force and the friction force:
F = Fgx - Ff.
We can substitute the expressions for Fgx, Fgy, and Ff into the equation for F:
F = (Fg * sinθ) - (μ * Fg * cosθ).
Since the sled moves with constant velocity, F = 0. Substituting this into the equation, we get:
0 = (Fg * sinθ) - (μ * Fg * cosθ).
Divide both sides of the equation by Fg:
0 = sinθ - (μ * cosθ).
Solve for μ:
μ = tanθ.
Now we can substitute the given angle θ = 60.0° into the equation to find the coefficient of friction μ:
μ = tan(60.0°).
Using a calculator, we find that tan(60.0°) is approximately 1.73.
Therefore, the coefficient of friction between the sled and the ice is approximately 1.73.