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

Answer 1

Given

Mass m = 44

Coefficient of static friction µ_s = 0.1

Coefficient of kinetic friction µ_k = 0.03

Angle of inclination θ = 25

Solution

A)

The normal force acting on the bag N = mgcosθ

= 44 x 9.8 x cos25

= 390.8 N

Force of Static friction F_s = µ_sN

= 0.1 x 390.8

= 39.08 N

The minimum force required F_min = 39.08 – mgsinθ

= 39.08 - 182.2

= -143.12 N

This means no external force is needed. The bag itself is moving downstairs

The kinetic friction F_­k = µ_kN

= 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 = mgcosθ

= 44 x 9.8 x cos25

= 390.8 N

Force of Static friction F_s = µ_sN

= 0.1 x 390.8

= 39.08 N

The minimum force required F_min = 39.08 + mgsinθ

= 39.08 + 182.2

= 221.28 N

The kinetic friction F_­k = µ_kN

= 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