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 E FI| = 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)

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 E FI| = 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)

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter5: More Applications Of Newton’s Laws

Section: Chapter Questions

Problem 1P

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A crate of mass 100.0 kg rests on a rough surface inclined at an angle of 37.0with the horizontal. A massless rope to which a force can be applied parallel to the surface is attached to the crate and leads to the top of the incline. In its present state, the crate is just ready to slip and start to move down the plane. The coefficient of friction is 80of that for the static case. (a) What is the coefficient of static friction? (b) What is the maximum force that can be applied upward along the plane on the rope and not move the block? (c) With a slightly greater applied force, the block will slide up the plane. Once it begins to move, what is its acceleration and what reduced force is necessary to keep it moving upward at constant speed? (d) If the block is given a slight nudge to get it started down the plane, what will be its acceleration in that direction? (e) Once the block begins to slide downward, what upward force on the rope is required to keep the block from accelerating downward?
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If an object is to rest o an incline without slipping, then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is =tan1s . You may use the result of the previous problem. Assume that a=0and that static friction has reached its maximum value.
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A physics major is cooking breakfast en she notices that the frictional force between her steel spatula and Teflon trying pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials, she quickly calculates the normal force. What is it?
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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