Part A A box with mass m is dragged across a level floor having a coefficient of

Question

Part A A box with mass m is dragged across a level floor having a coefficient of kinetic friction by a rope that is pulled upward at an angle above the horizontal with a force of magnitude F In terms of m.ilk: , and g, obtain an expression for the magnitude of force required to move the box with constant speed. Submit My Answers Give Up 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 Ha = 0.35. Use the result of part A to answer he instructors question. Submit My Answers Give Up

Explanation / Answer

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) As much as I love physics, I hope if EMTs ever have to drag me somewhere, they will not stop to determine the coefficient of friction...but back to physics.

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= 292.8 N

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