A 50-kg box is resting on a horizontal floor. A force of 250 N directed at an angle of 30.0° below the horizontal is applied to the box. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.30. What is the force of friction on the box?

f = 216.5 N

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=50 kg : mass of the box

F=250 N : Force applied to the box

θ = 30°: angle θ of the F=250N below the horizontal

μs=0.40: coefficient of static friction between the box and the surface

μk = 0.30: coefficient of kinetic friction between the box and the surface

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the floor and the y-axis in the direction perpendicular to it.

W: Weight of the box : In vertical direction downaward

N : Normal force : In vertical direction the upaward

F : Force applied to the box: In 30° below the horizontal direction.

f : Friction force: In horizontal direction

Calculated of the weight of the box

W = m*g = (50 kg) * (9.8 m/s²) = 490 N

x-y components F

Fx = Fcosθ = (250 N) * cos (30) ° = 216.5 N

Fy = Fsinθ = (250 N) * sin (30) ° = 125 N

Calculated of the Normal force

∑Fy = m*ay ay = 0

N-W-Fy = 0

N-490 N-125 N = 0

N = 615 N

Calculated of the static friction force (fs):

fs=μs*N = 0.4*615 N

fs = 246 N

Problem development

We have to

fs = 246 N

Fx = 216.5 N

Since the minimum force Fx is less than the fs, means that the box remains at rest and we apply Newton's first law to calculate f:

∑Fx = 0

f-Fx=0

f = Fx

f = 216.5 N