ENGG150 – Electrical and Mechanical Principles

Mechanical 1
ENGG150 ASSIGNMENT 2 (MECHANCIAL)
Student’s Full Name
Student Number
Tutor’s Name
Institution

Mechanical 2
Question 1
Let MA and FR be 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 )
1 Alys Holden et al, Structural Design For The Stage (CRC Press, 2nd ed, 2015)

Mechanical 3
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 = 0 3
(50)(3
5¿(4) + (50)(4
5 ¿(0.2) – (1000×9.81×10-3 kN)(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
2 Alys Holden et al, Structural Design For The Stage (CRC Press, 2nd ed, 2015)3 R. C. Hibbeler, Statics And Mechanics Of Materials (Pearson, 5th ed, 2018)