Question

Brian is trying to pull Julie on a sled across a flat, snowy field. Brian pulls on a rope attached to the sled. His pulling force is directed forward and upward at an angle of 30° above horizontal. Julie’s mass is 50 kg, and the sled’s mass is 8 kg. If the coefficient of static friction between the sled runners and the snow is 0.10, how much force must Brian exert on the rope to start moving the sled?

Answer 1

65,63 N

Step-by-step explanation:

Let F be the force with which Brian pulls the sled. Since the force is 30° above the horizontal, the component of the force in the forward direction is Fcos30°. This forward component equals the static frictional force, f for the sled to just begin to move.

So Fcos30° = f

Now f = μN where μ = coefficient of static friction between the sled and the snow = 0.10 and N = the normal force on the sled = weight of sled and Julie = (m + M)g where M = mass of Julie = 50 kg and m = mass of sled = 8 kg. g = acceleration due to gravity = 9.8 m/s².

Fcos30° = f = μN = μ(m + M)g

Fcos30° = μ(m + M)g

making F subject of the formula,

F = μ(m + M)g/cos30°

substituting the values of the variables, we have

F = 0.10(8 kg + 50 kg) × 9.8 m/s²/cos30°

F = 0.10 × 58 kg × 9.8 m/s²/cos30°

F = 56.84 N/0.8660

F = 65.63 N

So, Brian must exert a force of 65.63 N on the rope for the sled to start moving