Question

A box with mass m is dragged across a level floor having a coefficient of kinetic friction ?k by a rope that is pulled upward at an angle ? above the horizontal with a force of magnitude F.

Part A

In terms of m, ?k, ?, and g, obtain an expression for the magnitude of force required to move the box with constant speed.

Part B

Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a 90-kg patient across a floor at constant speed by pulling on him at an angle of 25? above the horizontal. By dragging some weights wrapped in an old pair of pants down the hall with a spring balance, you find that ?k=0.35. Use the result of part A to answer the instructor's question.

Answer 1

a) if the box slides with constant velocity, all the forces on the box sum to zero, so we break down all forces into the horizontal and vertical components

in the horizontal direction:

there is the horizontal component of the dragging force and the resistance of friction

horizontal component = F cos(theta)
friction force = u N where u is the coefficient of friction and N is the normal force (we do not assume N=weight)

vertical direction:

vertical component of pulling force = F sin(theta), the normal force acting up and the weight acting down

we have our two equations:

F cos(theta) = u N

F sin(theta) + N = mg

the second equation tells us that

N= mg-F sin (theta)

(the normal force is less than the weight since the dragging force has a vertical component that reduces the normal force)

use this equation for N in the horizontal equation above:

F cos(Theta) = u N

Fcos(theta) = u (mg - F sin(theta))

collect F terms and solve for F:

F = u mg /(cos(theta) + u sin(theta))

b)

we use the equation we derived above for the case:

m=90kg, theta = 25 deg, u = 0.35

F = 0.35x90kgx9.8m/s/s/[cos 25 +0.35 sin 25]

F= 308.87/1.10=280.79 N