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# ENGG150 – Electrical and Mechanical Principles

Added on - 05 Nov 2021

• ENGG150

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Mechanical1
ENGG150 ASSIGNMENT 2 (MECHANCIAL)
Student’s Full Name
Student Number
Tutor’s Name
Institution
Mechanical2
Question 1
LetMAandFRbe the moment at point A and resultant force1
Use equilibrium condition.
Fx= -30sin 30o+(26×5
13)
= - 15 + 10
= - 5 kN
Also total forces in y-direction is 0
Fy= - (30 cos 30o) -(26×12
13)
= - 25.9807 – 24
= - 49.9807kN
Calculating the resultant forces
FR=¿¿
=¿¿
= 50.2302
Getting the angle
tan θ =Fy
Fx
=(49.9807
5)
Θ = tan-1(49.9807
5)
1Alys Holden et al,Structural Design For The Stage(CRC Press, 2nd ed, 2015)
Mechanical3
θ = 84.28720
Getting moment at point A2
MA= (30 × sin 300× 0.3) – (30 cos300× 2) - 45 –(26×12
13×6)-(26×5
13×0.3)
= 4.5 – 51.96 – 45 - 144 – 3
= -239.46 kNm
Question 2
a)
b)
Mc= 03
(50)(3
5¿(4) + (50)(4
5¿(0.2) – (1000×9.81×10-3kN)(x) = 0
x = 13.04m
Fx= 0
Cx– (50) (4
5¿= 0
Cx= 40 kN
Fy= 0
-Cy+ (50) (3
5¿– (1000×9.81×10-3) = 0
2Alys Holden et al,Structural Design For The Stage(CRC Press, 2nd ed, 2015)3R. C. Hibbeler,Statics And Mechanics Of Materials(Pearson, 5th ed, 2018)

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