Question

The diagram below is a top-down view of two children pulling a 10.7-kg sled along the snow. The first child exerts a force of F₁ = 16 N at an angle θ₁ = 45° counterclockwise from the positive x direction. The second child exerts a force of F₂ = 6 N at an angle θ₂ = 30° clockwise from the positive x direction.

(a) Find the magnitude and direction of the friction force acting on the sled if it moves with constant velocity.

(b) What is the coefficient of kinetic friction between the sled and the ground?

‪(c) What is the magnitude of the acceleration of the sled if F₁ is doubled and F₂ is halved in magnitude?

Answer 1

F₁ = 16 N at an angle θ₁ = 45° counterclockwise from the positive x direction. = 16 cos 45 i + 16 sin 45 j

F₂ = 6 N at an angle θ₂ = 30° clockwise from the positive x direction = 6 cos 30 i - 6 sin 30 j

m = 10.7 Kg

(a) Find the magnitude and direction of the friction force acting on the sled if it moves with constant velocity.

As the body moves at constant velocity, Friction F = - ( 16 cos 45 i + 16 sin 45 j + 6 cos 30 i - 6 sin 30 j) = - (16.51 i + 8.31 j) N

Magnitude of friction =

Angle = tan^- (8.31/16.51) = 26.72 degree anticlockwise of - x axis (negative x axis)

(b) What is the coefficient of kinetic friction between the sled and the ground?

Coefficient of kinetic friction =

‪(c) What is the magnitude of the acceleration of the sled if F₁ is doubled and F₂ is halved in magnitude?

New forces are 2(16 cos 45 i + 16 sin 45 j) and 0.5(6 cos 30 i - 6 sin 30 j)

= (32 cos45 + 3cos30) i + (32 sin 45 - 3 sin 30) j = 25.225 i + 21.127j = 32.9 N

Friction = 18.49 N

So effective force = 32.9-18.49 = 14.41 N

Acceleration = 14.41/10.7 = 1.347 m/s²