The coefficient of friction between blocks is us = 0.4, at what angle will the block start sliding if it is initially at rest? Solution Let us obtain the net force in the axis perpendicular to the inclined plane. By virtue of the first law of motion (equilibrium), the net force is 2 F perpendicular =n - Fgperpendicular= Thus, the normal force is n = m Cos On the axis parallel to the inclined, the net force is 2 FI| - Fg|| -fs.max = Substituting the forces, we get the following: mg smg By simplifying, we get the following: tan(theta) = mus %3D

The coefficient of friction between blocks is us = 0.4, at what angle will the block start sliding if it is initially at rest? Solution Let us obtain the net force in the axis perpendicular to the inclined plane. By virtue of the first law of motion (equilibrium), the net force is 2 F perpendicular =n - Fgperpendicular= Thus, the normal force is n = m Cos On the axis parallel to the inclined, the net force is 2 FI| - Fg|| -fs.max = Substituting the forces, we get the following: mg smg By simplifying, we get the following: tan(theta) = mus %3D

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Description: The coefficient of friction between blocks is us = 0.4, at what angle will the block start sliding if it is initially at rest?SolutionLet us obtain the net force in the axis perpendicular to the inclined plane.By virtue of the first law of motion (e

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter5: Newton's Law Of Motion

Section: Chapter Questions

Problem 5.9CYU: Check Your Understanding One end of a 3.0-m rope is tied to a tree; the other, end is tied to a car...

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Problem The coefficient of friction between blocks is us = 0.4, at what angle will the block start sliding if it is initially at rest? Solution Let us obtain the net force in the axis perpendicular to the inclined plane. By virtue of the first law of motion (equilibrium), the net force is begin inline style sum from blank to blank of end styleF perpendicular = n - Fgperpendicular = Thus, the normal force is n = mcos On the axis parallel to the inclined, the net force is begin inline style sum from blank to blank of end styleF|| = Fg|| - fs,max = Substituting the forces, we get the following: mg - smg = By simplifying, we get the following: tan(theta) = mus Thus, the angle that the object will slide is: (rounded of to the ones)
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