Lesson_2.3_Homework_Solution

Lesson_2.3_Homework_Solution - Lesson 8 Homework Solutions...

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Lesson 8 Homework Solutions 1. A 50-kg crate is at rest on a level surface. If the coefficient of static friction between the crate and the surface is 0.72, what horizontal force is required to move the crate? The horizontal force required to move the crate F = f s , the force of static friction. The weight of the crate = 490 N acts vertically down and the normal force F n = 490 N f s = µ k F n = 0.72 × 490 = 353 N f s 490 N F 50 kg Fn = 490 N 2. A horizontal force of 40 N is required to move a 10-kg crate on a level surface. (i) What is the coefficient of static friction between the surface and the crate? (ii) Once the crate is set in motion, only a force of 35 N is required to keep it moving uniformly. What is the coefficient of kinetic friction between the two surfaces? (i) Since a force of 40 N is required to overcome the force of static friction, frictional f s = 40 N The normal force F n = mg = 10 kg × 9.8 ms -2 = 98 N f s = µ s F n 40 N 98 41 N 0. s s n f F µ = == (ii) 35 N will keep it moving means that the force of kinetic friction is: f k = 35 N f k = µ k F n 35 N 98 36 N 0. k k n f F = 3. A 10-kg block is placed on an inclined plane of inclination 37 o . (i) What is the force pulling the block down the plane? (ii) Ignoring friction, what is the minimum force required to pull the block up the plane without acceleration? (iii) If the angle of inclination is increased
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to 42 o , the block begins to slide down without acceleration. What is the coefficient of kinetic friction between the block and the surface? The figure on the right, if there is no friction between the inclined plane and 3.0 kg mass, find the acceleration of the masses and the tension on the string. 10 kg 37 o 5 kg 59 N T 49 N 98 N The weight of the 10-kg mass = 10 kg × 9.8 ms -2 = 98 N acts vertically down. (i) The component of the weight mg sin θ = 98 sin 37 o = 59 acts down the plane trying to pull the block down. (ii) Since friction is ignored, this is the amount of force that is required to pull the block up the plane without acceleration. (iii) When the angle is 42 o , the force pulling the block down the plane = 98 sin 42 o = 65.6 N. When the block begins to slide down without acceleration, this force is balanced by the force of kinetic friction f k up the plane.
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Lesson_2.3_Homework_Solution - Lesson 8 Homework Solutions...

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