Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 10)
10.
If the resultant of two equal forces has the same magnitude as either of the forces, then the angle between the two forces is
Discussion:
35 comments Page 1 of 4.
Kesav said:
4 years ago
According to Lamis theorem, sine α = sine β = sine γ.
i.e. α + β + γ =360 so 360 /3 =120 =α = β= γ.
i.e. α + β + γ =360 so 360 /3 =120 =α = β= γ.
(29)
Panneerselvam N said:
4 years ago
To find the resultant
R= √(A2)+(B2)+2ABcosx
As given in the query, the magnitude of the force is the same and let it be f whose resultant be f too.
Substituting the variables in the above equation we get
f= √(f2)+(f2)+2(f2)cosx
Squaring on both sides we get
f2=2(f2)+2(f2)cosx
f2/f2 = 2 + 2 cos x
1 - 2 = 2 cosx
-1/2 = cos x
Therefore x=120°.
R= √(A2)+(B2)+2ABcosx
As given in the query, the magnitude of the force is the same and let it be f whose resultant be f too.
Substituting the variables in the above equation we get
f= √(f2)+(f2)+2(f2)cosx
Squaring on both sides we get
f2=2(f2)+2(f2)cosx
f2/f2 = 2 + 2 cos x
1 - 2 = 2 cosx
-1/2 = cos x
Therefore x=120°.
(12)
Kiran kumar said:
4 years ago
According to Lamis theorem : p/sinα+q/sinβ+r/sinγ.
S V Manikandan. said:
5 years ago
The two forces acting with the same magnitude and resultant too same magnitude. So the angle between each two forces is equal. So we can divide the circle into three equal segments. The angle of a circle is 360°. Then 360/3 = 120°each.
(2)
Vaibhav said:
5 years ago
The resultant force is the same value of one of this either forces, so we can assume R=P, P1=P, P2=P.
Let using parallelogram therom,
R^2 = P1^2+P2^2+2P1P2COSθ
P^2 = P^2+P^2+2P^2COSθ
P^2 = 2P^2+2P^2COSθ
2P^2COSθ = -P^2
And θ=120°.
Let using parallelogram therom,
R^2 = P1^2+P2^2+2P1P2COSθ
P^2 = P^2+P^2+2P^2COSθ
P^2 = 2P^2+2P^2COSθ
2P^2COSθ = -P^2
And θ=120°.
(3)
Ranga... said:
7 years ago
Two force and one resultant.
Than 360/3-120.
Than 360/3-120.
Sachin said:
7 years ago
Based on the Law of parallelogram.
(1)
Naiya said:
8 years ago
I think we can apply Lami's theorem.
ABE said:
8 years ago
R= √(A^2+B^2+2ABcosθ).
Here A=F,B=F ;We need R=F
squaring on bhs.
F^2 =2F^2+2F^2cosα
on solving,
-F^2=2F^2cosα.
cosα=-0.5.
α=120 DEGREE.
Here A=F,B=F ;We need R=F
squaring on bhs.
F^2 =2F^2+2F^2cosα
on solving,
-F^2=2F^2cosα.
cosα=-0.5.
α=120 DEGREE.
Ajay said:
8 years ago
Can anyone tell exact explanation?
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