Assume w = 9 kn/m. Part a determine the magnitude of the resultant force
Answers
Given
Weight density (w) = 9 kN/m
Since the load is uniformly varying, therefore the resultant of the load that is left to point "A" is
P
1
=
0.5
∗
w
∗
3
P
1
=
0.5
∗
9
∗
3
P
1
=
13.5
k
N
Now, the point of application of the resultant load from point A would be
x
=
1
3
∗
3
x
=
1
m
Now, the resultant of the load that is right to point A
P
2
=
0.5
∗
w
∗
6
P
2
=
0.5
∗
9
∗
6
P
2
=
27
k
N
Now, the point of application of the resultant load would be
x
=
1
3
∗
6
x
=
2
m
(a)
Now, the net resultant force
P
=
P
1
+
P
2
P
=
13.5
+
27
P
=
40.5
k
N
(b)
Now, moment about point A would be zero, therefore
∑
M
A
=
0
−
(
P
1
∗
1
)
+
(
P
2
∗
2
)
−
(
R
B
∗
6
)
=
0
R
B
∗
6
=
(
27
∗
2
)
−
(
13.5
∗
1
)
R
B
=
6.75
k
N
Now, if we replace the P1 and P2 by the resultant force that is "P", therefore
∑
M
A
=
0
P
∗
x
−
R
B
∗
6
=
0
40.5
∗
x
=
6.75
∗
6
x
=
1
m
Therefore net resultant force would be at 1 m right from point "A"