Contents
Task 1..........................................................................................................................................................1
Part(a).....................................................................................................................................................1
Part b.......................................................................................................................................................2
Task 2..........................................................................................................................................................3
Task 3..........................................................................................................................................................8
Task 4........................................................................................................................................................10
Part a.....................................................................................................................................................10
Part b.....................................................................................................................................................11
Task 5........................................................................................................................................................12
Part a.....................................................................................................................................................12
Part b.....................................................................................................................................................13
Task 1
Part(a)
From Table 1,
W 1=3.4 kN=3,400 N
820 N
w 2= m
L1=3.5 m.
, Free body diagram
Since the system is at equilibrium then
R∑+ R −34 00 N −
F=0 820 N
× 3 m =0
A B ( m )
RA + RB=5,860 N .
Taking moments at A and considering the system is at equilibrium
∑ M clockwise = ∑ M anticlockwise
(3400 N × 3.5 m) 820 N ×3 m× 6.5 m
+ =R ×9.5 m
m ( ) B
11900 Nm+15990 Nm=9.5 m× RB
278990 N / m
R= =2,935.79 N
B
9 ⋅5 m
Therefore,
R A +2935.79 N=5,860 N
R A =5,860 N −2935.79 N =2,924.21 N .
Therefore, the reactions at A and B are 2,924N and 2,936N respectively
Part b
From Table 1,
Task 1..........................................................................................................................................................1
Part(a).....................................................................................................................................................1
Part b.......................................................................................................................................................2
Task 2..........................................................................................................................................................3
Task 3..........................................................................................................................................................8
Task 4........................................................................................................................................................10
Part a.....................................................................................................................................................10
Part b.....................................................................................................................................................11
Task 5........................................................................................................................................................12
Part a.....................................................................................................................................................12
Part b.....................................................................................................................................................13
Task 1
Part(a)
From Table 1,
W 1=3.4 kN=3,400 N
820 N
w 2= m
L1=3.5 m.
, Free body diagram
Since the system is at equilibrium then
R∑+ R −34 00 N −
F=0 820 N
× 3 m =0
A B ( m )
RA + RB=5,860 N .
Taking moments at A and considering the system is at equilibrium
∑ M clockwise = ∑ M anticlockwise
(3400 N × 3.5 m) 820 N ×3 m× 6.5 m
+ =R ×9.5 m
m ( ) B
11900 Nm+15990 Nm=9.5 m× RB
278990 N / m
R= =2,935.79 N
B
9 ⋅5 m
Therefore,
R A +2935.79 N=5,860 N
R A =5,860 N −2935.79 N =2,924.21 N .
Therefore, the reactions at A and B are 2,924N and 2,936N respectively
Part b
From Table 1,
