derive an expression for magnitude and direction of resultant of two forces acting at a point with an acute angle
Answers
Explanation:
Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure.
Let θ be the angle between P and Q and R be the resultant vector. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.
So, we have
R = P + Q
Now, expand A to C and draw BC perpendicular to OC.
From triangle OCB,
OB
2
=C
2
+BC
2
or OB
2
=(OA+AC)
2
+BC
2
. . . . . . ( i )
Intriangle ABC,
cosθ=
AB
AC
or , AC = AB cosθ
or , AC = OD cosθ
= Q cosθ [ AB = OD = Q ]
Also,
cosθ=
AB
BC
or , BC = AB sinθ
or , BC = OD sinθ
= Q sinθ [ AB = OD = Q }
Magnitude of resultant:
Substituting value of AC and BC in ( i ), we get
R
2
=(P+Qcosθ)
2
+(Qsinθ)
2
or , R
2
=P
2
+2PQcosθ+Q
2
cos
2
θ+Q
2
sin
2
θ
or , R
2
=P
2
+2PQcosθ+Q
2
R =
P
2
+2PQcosθ+Q
2
Which is the magnitude of resultant.
Direction of resultant :
Let ϕ be the angle made by resultant R with P . Then,
From triangle OBC,
tanϕ=
OC
BC
=
OA+AC
BC
or , tanϕ=
P+Qcosθ
Qsinθ
ϕ=tan
−1
(
P+Qcosθ
Qsinθ
)
which is the direction of resultant.
solution
Answer:
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Explanation:
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