Part 1: A student decides to move a box of books into her dormitory room by pull

Question

Part 1: A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 170 N at an angle of 30 degrees above the horizontal. The box has a mass of 26.9 kg, and the coefficient of friction between box and floor is .271.
The acceleration of gravity is 9.8 m/s.
Find the acceleration of the box.

Part 2: The student now starts moving the box up a 13.6 degree incline, keeping her 170 N force directed at 30 degrees above the line of the incline.
If the coefficient of friction is unchanged, what is the new acceleration of the box?

Explanation / Answer

horizontal component of force = 170*cos(30) = 147.25 N vertical component is = 170 sin(30 = 84.96 N normal force on the box = (m*g)-(vertical component of force by student) = (26.9)*(9.8) - 84.96 = 178.66 so frictional force = 0.271*178.66 = 48.42 N total force in horizontal direction = horizontal component - frictional force = 147.25 - 48.42 = 98.83 N accleartion = 98.83/26.9 = 3.67 m/s^2 b) now he is moving up an incline of angle = 13.6 degrees so force in the line of incline by student= 170*cos(30) = 147.25 N force vertical to the line of incline by student is = 170 sin(30 = 84.96 N normal force on the box = (m*g*cos(13.6))-(vertical component of force by student) = (26.9)*(9.8)*cos(13.6) - 84.96 = 171.276 N so frictional force = 0.271*171.276 = 46.416 N negative force along the incline due to weight = m*g*sin(13.6) = 61.96 N total force in horizontal direction = horizontal component - frictional force-(weight component) = 147.25 - 46.416 -61.96 = 38.874 N accleartion = 38.87400/26.9 = 1.445 m/s^2

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