w 750 N + 0.91 # = 0. By solving and Ⓒ F=-177-7 N N = 1400-69 N Static friction force F₁ = H₂ N D

Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN:9780078807213
Author:Paul W. Zitzewitz
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Chapter7: Gravitation
Section: Chapter Questions
Problem 73A
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What is the solution to get the answer for the highlighted (yellow) equation

Given
The angle of enclined surface with Horizontal (0) = 25°
Applied force (P)
Static coff of friction (Mg) = 0.35
Kinetic coff of friction (UK) = 0.25
.
= 750N
Transcribed Image Text:Given The angle of enclined surface with Horizontal (0) = 25° Applied force (P) Static coff of friction (Mg) = 0.35 Kinetic coff of friction (UK) = 0.25 . = 750N
S
FB-D
Now,
1200 N
Now
Balancing Horizontal force
Balancing force perpendicular to Inclined plane.
(Vertical Force)
N Co 25° + f sin 25° = 1260 N
(0.91) N + (0.42) F = 1260 N
By solving and
IN
750 N+FOD 25° = Nsin 25°
750 N + 0.91
= 0.42 N
F = -177-7 N
N = 1400-69 N
Static friction force.
:: F <F₂ }
-0
F₁ = μg N
-
= 0.35 x 1450.69 N
= 490.24 N
N = Normal force
F = friction force.
equilibrium.
and i't act downward.
disn. Lee act dirn opposite to assumed dish along
inclined plane)
so the Block is in
Itence Mag of friction force is 177.7 N
Transcribed Image Text:S FB-D Now, 1200 N Now Balancing Horizontal force Balancing force perpendicular to Inclined plane. (Vertical Force) N Co 25° + f sin 25° = 1260 N (0.91) N + (0.42) F = 1260 N By solving and IN 750 N+FOD 25° = Nsin 25° 750 N + 0.91 = 0.42 N F = -177-7 N N = 1400-69 N Static friction force. :: F <F₂ } -0 F₁ = μg N - = 0.35 x 1450.69 N = 490.24 N N = Normal force F = friction force. equilibrium. and i't act downward. disn. Lee act dirn opposite to assumed dish along inclined plane) so the Block is in Itence Mag of friction force is 177.7 N
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