A 25-kg package, initially at rest, is being pushed to the right by applying a force P on corner A in the direction shown in the figure. At the instant shown, corner C is already 0.250 m to the right of edge E of the platform. Assume that the coefficients of non-sliding (static) and sliding (kinetic) friction between the package and the platform are μ = 0.40 and μ = 0.25, respectively. 0.900 m D 0.600 m ► 0.250 m
A 25-kg package, initially at rest, is being pushed to the right by applying a force P on corner A in the direction shown in the figure. At the instant shown, corner C is already 0.250 m to the right of edge E of the platform. Assume that the coefficients of non-sliding (static) and sliding (kinetic) friction between the package and the platform are μ = 0.40 and μ = 0.25, respectively. 0.900 m D 0.600 m ► 0.250 m
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
Given the following, answer the 2 subproblems

Transcribed Image Text:A 25-kg package, initially at rest, is being pushed to the right by applying a force P on corner A in the direction shown in the figure. At the instant shown,
corner C is already 0.250 m to the right of edge E of the platform. Assume that the coefficients of non-sliding (static) and sliding (kinetic) friction between
the package and the platform are μ = 0.40 and μ = 0.25, respectively.
0.900 m
20⁰
0.600 m
0.250 m

Transcribed Image Text:1. Which of the following gives the most ideally constructed force diagram of the package? Note: In this diagram, N is the single equivalent force that
ideally replaces the effect of contact surface DE to the package.
A.
C₁
A.
B.
10.450
D.
0.900 m
19.450 m
0.600 m
0.600 m
0.250m
B. P
D.
0.900m
10.450 m
W
0.900 m
10.450 m
0.600 m
2. Which of the following sets of equilibrium equations for the package is CORRECT?
EF: P cos 20-f=0
ΣFy: P sin 20-W+N = 0
Mc:-(P cos 20) (0.6) (P sin 20) (0.9) + W (0.45) Nx = 0
0.600 m
0.250 m
ΣFx:P sin 20+ f = 0
EF: P cos 20-W+N = 0
Mc:-(P sin 20) (0.6) (P cos 20) (0.9) + W (0.45) Nx = 0
EF:P cos 20 - f = 0
EF: P sin 20-W+N=0
Mc:-(P cos 20) (0.6) (P sin 20) (0.9) + W (0.45)-N (0.25) = 0
EF:P sin 20 + f = 0
ΣFy: P cos 20-W+N=0
Mc:-(P sin 20) (0.6) + (P cos 20) (0.9) + W (0.45)-N (0.25) = 0
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