Chapter 14

# Chapter 14 - Engineering Mechanics Dynamics Chapter 14...

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Unformatted text preview: Engineering Mechanics - Dynamics Chapter 14 Problem 14-1 A woman having a mass M stands in an elevator which has a downward acceleration a starting from rest. Determine the work done by her weight and the work of the normal force which the floor exerts on her when the elevator descends a distance s . Explain why the work of these forces is different. Units Used: kJ 10 3 J Given: M 70 kg g 9.81 m s 2 a 4 m s 2 s 6 m Solution: M g N p M a N p M g M a N p 406.7 N U W M g s U W 4.12 kJ U NP s N p U NP 2.44 kJ The difference accounts for a change in kinetic energy. Problem 14-2 The crate of weight W has a velocity v A when it is at A . Determine its velocity after it slides down the plane to s = s' . The coefficient of kinetic friction between the crate and the plane is P k . Given: W 20 lb a 3 b 4 v A 12 ft s s' 6 ft P k 0.2 Solution: T atan a b § © · ¹ N C W cos T + , F P k N C Guess v' 1 m s Given 1 2 W g § © · ¹ v A 2 W sin T + , s' . F s' 1 2 W g § © · ¹ v' 2 v' Find v' ( ) v' 17.72 ft s 242 Engineering Mechanics - Dynamics Chapter 14 Problem 14-3 The crate of mass M is subjected to a force having a constant direction and a magnitude F , where s is measured in meters. When s = s 1 , the crate is moving to the right with a speed v 1 . Determine its speed when s = s 2 . The coefficient of kinetic friction between the crate and the ground is P k . Given: M 20 kg F 100 N s 1 4 m T 30 deg v 1 8 m s a 1 s 2 25 m b 1 m 1 P k 0.25 Solution: Equation of motion: Since the crate slides, the friction force developed between the crate and its contact surface is F f = P k N N F sin T + , . M g N M g F sin T + , Principle of work and Energy: The horizontal component of force F which acts in the direction of displacement does positive work, whereas the friction force F f P k M g F sin T + , + , does negative work since it acts in the opposite direction to that of displacement. The normal reaction N , the vertical component of force F and the weight of the crate do not displace hence do no work. F cos T + , P k N M a F cos T + , P k M g F sin T + , + , M a a F cos T + , P k M g F sin T + , + , M a 2.503 m s 2 v d v d s a v 2 2 v 1 2 2 a s 2 s 1 + , . v 2 v 1 2 2 a s 2 s 1 + , . ª « ¬ º » ¼ v 13.004 m s 243 Engineering Mechanics - Dynamics Chapter 14 * Problem 14-4 The “ air spring ” A is used to protect the support structure B and prevent damage to the conveyor-belt tensioning weight C in the event of a belt failure D . The force developed by the spring as a function of its deflection is shown by the graph. If the weight is W and it is suspended a height d above the top of the spring, determine the maximum deformation of the spring in the event the conveyor belt fails. Neglect the mass of the pulley and belt....
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Chapter 14 - Engineering Mechanics Dynamics Chapter 14...

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