Your neighbor is pushing a 44.0-kg bag of dirty laundry across a parking lot as
ID: 1328652 • Letter: Y
Question
Your neighbor is pushing a 44.0-kg bag of dirty laundry across a parking lot as shown in the figure below. The coefficient of static friction between the bag and the asphalt is 0.1, and the coefficient of kinetic friction is 0.03. Assume that up and right are the positive directions.Determine the following for case (a), where he's pushing downward at an angle of 25°. (a) Calculate the minimum force F he must exert to get the stinky bag moving. Fmin = N (b) What is its acceleration once it starts to move, if that force is maintained? a = m/s2 PopUp: Force Equations 2. Determine the following for case (b), where he's pulling upward at an angle of 25 °. (a) Calculate the minimum force F he must exert to get the stinky bag moving. Fmin = N (b) What is its acceleration once it starts to move, if that force is maintained? a = m/s2
Explanation / Answer
Given
Mass m = 44
Coefficient of static friction µs = 0.1
Coefficient of kinetic friction µk = 0.03
Angle of inclination = 25
Solution
A)
The normal force acting on the bag N = mgcos
= 44 x 9.8 x cos25
= 390.8 N
Force of Static friction Fs = µsN
= 0.1 x 390.8
= 39.08 N
The minimum force required Fmin = 39.08 – mgsin
= 39.08 - 182.2
= -143.12 N
This means no external force is needed. The bag itself is moving downstairs
The kinetic friction Fk = µkN
= 0.03 x 390.8N
= 11.724 N
This opposes motion during motion also the static friction vanishes once its start to move
so the resultant Force F = 143.12 - 11.724
= 131.396 N
Acceleration a= F/m = 2.99 m/s2
B)
The normal force acting on the bag N = mgcos
= 44 x 9.8 x cos25
= 390.8 N
Force of Static friction Fs = µsN
= 0.1 x 390.8
= 39.08 N
The minimum force required Fmin = 39.08 + mgsin
= 39.08 + 182.2
= 221.28 N
The kinetic friction Fk = µkN
= 0.03 x 390.8N
= 11.724 N
This opposes motion during motion so the resultant Force
F = 221.28 – 11.724 -182.2
= 27.356
Acceleration a= 27.356 / 44 = 0.622m/s2
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