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Subject: Dynamics of Rigid Bodies Topic: "Work – Energy Relation, Power and Efficiency." Kindly solve the activty no. 1 with the same answer provided and with complete solution and FBD.

Subject: Dynamics of Rigid Bodies Topic: "Work – Energy Relation, Power and Efficiency." Kindly solve the activty no. 1 with the same answer provided and with complete solution and FBD.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Subject: Dynamics of Rigid Bodies

Topic: "Work – Energy Relation, Power and Efficiency."

Kindly solve the activty no. 1 with the same answer provided and with complete solution and FBD.

 

Activity 1. In Problem No. 2, solve for v, when b.)S = 5.50 m. (Ans. v=3.48 m/s)
Transcribed Image Text:Activity 1. In Problem No. 2, solve for v, when b.)S = 5.50 m. (Ans. v=3.48 m/s)
EXAMPLE 2. A body weighing 2225 N is being dragged along a rough horizontal plane by a force "F" = 445 N. If the coefficient of friction is 1/12 and the line of the pull makes an angle of 18° above the horizontal, what are the velocities required from rest in the first 3.65 m and in the first 5.50 m? W=2225N P = 445N 180 F=µN Solution: EFv=0; + N-W + 445 sin 18° = 0 N = 2087.49 N But : F = 1/12 N = 2087.49/12 F= 173.96 N a. For S = 3.65 m IKE = initial kinetic energy = 0 PU = positive work = PS = P cose = 445 cos 18° (3.65) = 1544.75N.m NU = negative work = FS = 173.96(3.65) = 52.91 N.m FKE = final kinetic energy = ½ mv = ½ wv²lg = 2225(v²)/19.60 = 113.52² N m/s? Using the Work-Energy Relation Formula: PU – NU = FKE - IKE 1544.75 N.m – 52.91 N.m = 113.52 vN -0 m/s² 1544.75 kg.m/s² (m)– 52.91 kg.m/s² (m) = 113.52 v ka.m/s? m/s² V = 2.83 m/sans
Transcribed Image Text:EXAMPLE 2. A body weighing 2225 N is being dragged along a rough horizontal plane by a force "F" = 445 N. If the coefficient of friction is 1/12 and the line of the pull makes an angle of 18° above the horizontal, what are the velocities required from rest in the first 3.65 m and in the first 5.50 m? W=2225N P = 445N 180 F=µN Solution: EFv=0; + N-W + 445 sin 18° = 0 N = 2087.49 N But : F = 1/12 N = 2087.49/12 F= 173.96 N a. For S = 3.65 m IKE = initial kinetic energy = 0 PU = positive work = PS = P cose = 445 cos 18° (3.65) = 1544.75N.m NU = negative work = FS = 173.96(3.65) = 52.91 N.m FKE = final kinetic energy = ½ mv = ½ wv²lg = 2225(v²)/19.60 = 113.52² N m/s? Using the Work-Energy Relation Formula: PU – NU = FKE - IKE 1544.75 N.m – 52.91 N.m = 113.52 vN -0 m/s² 1544.75 kg.m/s² (m)– 52.91 kg.m/s² (m) = 113.52 v ka.m/s? m/s² V = 2.83 m/sans
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