The lumberjack pulls on the sled with 40 N at an angle of 30 degrees, pulling so the sled moves at a constant velocity. 1) What is the x component of the applied force? 2) What is the y component of the applied force? 3) If the loaded sled has a mass of 65 kg, what is the magnitude of the force of gravity? 4) What is the magnitude of the normal force acting on the sled? 5) What is the coefficient of friction between the snow and the sled?

1) (40 N) cos(30°) ≈ 34.6 N

2) (40 N) sin(30°) = 20 N

3) (65 kg) g = (65 kg) (9.80 m/s²) = 585 N

4) The net force on the sled acting in the vertical direction is made up of

• the sled's weight, 585 N, pointing downward

• the vertical component of the applied force, 20 N, pointing upward

• the normal force, with magnitude n, also pointing upward

The sled does not move up or down, so by Newton's second law,

∑ F = n + 20 N - 585 N = 0   ==>   n = 565 N

5) The net force in the horizontal direction consists of

• the horizontal component of the applied force, 34.6 N, acting in the direction the sled's movement (call this the positive direction)

• kinetic friction, with magnitude f, pointing in the opposite and negative direction

By Newton's second law,

∑ F = 34.6 N - f = 0   ==>   f ≈ 34.6 N

Now if µ is the coefficient of kinetic friction, then

f = µn   ==>   µ = f/n = (34.6 N) / (565 N) ≈ 0.0613