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JEE Main 27-Aug-2021 Shift 1 Solved Paper
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© examsnet.com
Question : 5
Total: 90
The resultant of these forces
OP, OQ, OR, OS
and
OT
is approximately ...... N.
[Take,
√
3
=
1.7
,
√
2
=
1.4
and given
^
i
and
^
j
unit vectors along X, Y axis
[27 Aug 2021 Shift 1]
9.25
^
i
+
5
^
j
3
^
i
+
15
^
j
2.5
^
i
−
14.5
^
j
−
1.5
^
i
−
15.5
^
j
Validate
Solution:
From given figure, we can write all the forces in vector notation as
OP
=
(
20
sin
30
°
^
i
+
20
cos
30
°
^
j
)
N
=
(
10
^
i
+
17.32
^
j
)
N
OQ
=
(
10
cos
30
°
^
i
+
10
sin
30
°
^
j
)
N
=
(
8.66
^
i
+
5
^
j
)
N
OR
=
(
20
cos
45
°
^
i
−
20
sin
45
°
^
j
)
N
=
(
14.14
^
i
−
14.14
^
j
)
N
OS
=
(
−
15
cos
45
°
^
i
−
15
sin
45
°
^
j
)
N
=
(
−
10.60
^
i
−
10.60
^
j
)
N
OT
=
(
−
15
sin
60
°
^
i
+
15
cos
60
°
^
j
)
N
=
(
−
12.99
^
i
+
7.5
^
j
)
N
The resultant of these forces is given as
F = OP + OQ + OR + OS + OT
F
=
[
(
10
^
i
+
17.32
^
j
)
+
(
8.66
^
i
+
5
^
j
)
+
(
14.14
^
i
−
14.14
^
i
)
+
(
−
10.60
^
i
−
10.60
^
j
)
+
(
−
12.99
^
i
+
7.5
^
j
)
]
F
=
9.21
^
i
+
5.08
^
j
N
F
≈
9.25
^
i
+
5
^
j
N
Thus, the resultant force is
F
≈
9.25
^
i
+
5
^
j
N
© examsnet.com
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