The Grizwald’s have just chopped down a Christmas tree and have decided to pull it out of the woods on a sled over the snow. Clark has attached a rope to the sled and is pulling the tree/sled system (mass = 22 kg) with a force of 50.0 N at an angle of 32˚ above the horizontal. The tree/sled is moving at constant speed. (a) Draw a force diagram of the situation. (b) Create qualitative vertical and horizontal equations for this situation (one each). (c) What is the coefficient of kinetic friction between the snow and the sled?
Force diagram:
Fn pointing up
Fg pointing down
Fa pointing right/upwards at 32 deg. to horizontal
Fk pointing left
Vertical forces equation (up is positive):
Sled is not moving up or down so:
Fn + Fg + Fa = 0
Fn + (-9.81)(22) + 50 sin 32 = 0
Use the above to solve for Fn.
Horizontal forces equation:
Horizontal velocity is constant so:
Fa + Fk =0
50cos32 + (-uk)(Fn) =0
Plug in Fn that you found above and solve for uk, the coefficient of kinetic friction.
(a) To draw a force diagram of the situation, we need to identify the forces acting on the tree/sled system.
1. Weight (W): This is the force acting downwards due to the gravitational pull on the tree/sled system. Its magnitude can be calculated using the formula: W = m * g, where m is the mass (22 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. Normal Force (N): This is the force exerted by the snow on the sled perpendicular to the surface. Since the sled is not sinking into the snow, the magnitude of the normal force is equal to the weight (N = W).
3. Tension Force (T): This is the force exerted by Clark on the rope, pulling the tree/sled system. Its magnitude is given as 50.0 N at an angle of 32˚ above the horizontal.
4. Friction Force (Ff): This is the force opposing the motion of the tree/sled system. Since the system is moving at constant speed, the friction force must be equal in magnitude and opposite in direction to the tension force (Ff = -T).
Arrow diagrams can be used to represent these forces, with the lengths of the arrows indicating the relative magnitudes of the forces. The weight (W) arrow should be pointing downwards, the normal force (N) arrow should be pointing upwards, the tension force (T) arrow should be pointing upwards and at an angle of 32˚, and the friction force (Ff) arrow should be pointing to the left.
(b) To create qualitative vertical and horizontal equations for this situation, we need to consider the forces acting in those directions.
Vertical Equation:
Since the tree/sled system is moving at constant speed, the vertical forces must be balanced. Therefore, the sum of the vertical forces is zero.
ΣFy = N - W = 0, where N is the normal force and W is the weight.
Horizontal Equation:
Again, since the system is moving at constant speed, the horizontal forces must also be balanced.
ΣFx = T + Ff = 0, where T is the tension force and Ff is the friction force.
(c) To find the coefficient of kinetic friction between the snow and the sled, we need to use the horizontal equation and the given information.
ΣFx = T + Ff = 0
50.0 N + Ff = 0
Since the tree/sled system is moving at a constant speed, the friction force is equal in magnitude and opposite in direction to the tension force.
Ff = -T = -50.0 N
The friction force can also be expressed as:
Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.
Substituting the known values:
-50.0 N = μ * W
Since N = W, we can substitute W for N:
-50.0 N = μ * m * g
We know the mass (m = 22 kg) and the acceleration due to gravity (g ≈ 9.8 m/s^2), so we can solve for the coefficient of kinetic friction (μ):
μ = (-50.0 N) / (22 kg * 9.8 m/s^2)