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

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

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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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