Beatrice (of example 3.8 in our thick textbook) is dragging a 41.8 lbf suitcase

Question

Beatrice (of example 3.8 in our thick textbook) is dragging a 41.8 lbf suitcase across the floor by the means of a rope. Starting from rest, she managed to give the suitcase an acceleration of 0.4 m/s2 to get it going, by pulling on the rope with a force of 92.5 N at a 27.3o angle up relative to the horizontal floor. What is the coefficient of friction between the floor and the suitcase? Beatrice (of example 3.8 in our thick textbook) is dragging a 41.8 lbf suitcase across the floor by the means of a rope. Starting from rest, she managed to give the suitcase an acceleration of 0.4 m/s2 to get it going, by pulling on the rope with a force of 92.5 N at a 27.3o angle up relative to the horizontal floor. What is the coefficient of friction between the floor and the suitcase? Beatrice (of example 3.8 in our thick textbook) is dragging a 41.8 lbf suitcase across the floor by the means of a rope. Starting from rest, she managed to give the suitcase an acceleration of 0.4 m/s2 to get it going, by pulling on the rope with a force of 92.5 N at a 27.3o angle up relative to the horizontal floor. What is the coefficient of friction between the floor and the suitcase?

Explanation / Answer

Here ,

mass of suitcase , m = 41.8 lbf

m = 19 Kg

acceleration , a = 0.4 m/s^2

Force , F = 92.5 N at 27.3 degree

let the coefficient of friction is u

Now , normal reaction , N = mg - F * sin(27.3)

N = 19 * 9.8 - 92.5 * sin(27.3)

N = 143.8 N

using second law of motion

19 * 0.4 = 92.5 * cos(27.3) - 143.8 * u

u = 0.522

the coefficient of friction between the floor and the suitcase is 0.522

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