Question #18135

A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 176 N at an angle of 24.4◦ above the horizontal. The box has a mass of 33.4 kg, and μk between the box and the floor is 0.22.
The acceleration of gravity is 9.81 m/s2 . Find the acceleration of the box. Answer in units of m/s2

Expert's answer

Let us find horizontal net force.

Horizontal student's force is 176N*cos(24.4) = 160.28 N

Then vertical student's force is sqrt(176^2-160.28^2)=72.2N

Then friction force is μk*(mg-72.2N) = 0.22(33.4*9.8 N - 72.2N) = 56.1 N

Then acceleration is net force/mass and is equal to

(160.28N-56.1N)/33.4 = 3.2 m/s^2

Horizontal student's force is 176N*cos(24.4) = 160.28 N

Then vertical student's force is sqrt(176^2-160.28^2)=72.2N

Then friction force is μk*(mg-72.2N) = 0.22(33.4*9.8 N - 72.2N) = 56.1 N

Then acceleration is net force/mass and is equal to

(160.28N-56.1N)/33.4 = 3.2 m/s^2

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