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 3 of 4.
Shams said:
10 years ago
The angle is 60 and not 120. It can be clearly understood from equilateral triangle and total angle in a triangle. And also from the cosine rule: A^2 = B^2+C^2-2BC cosA.
Tushar Chakraborty said:
10 years ago
Law of Cosine- R^2=P^2+Q^2-2PQ Cosθ
So θ = 60°
When Forces are Nose to tail, θ = 60°
When Forces are Tail to Tail, θ = 180°-60° = 120°
So θ = 60°
When Forces are Nose to tail, θ = 60°
When Forces are Tail to Tail, θ = 180°-60° = 120°
Abdul Khaleque said:
10 years ago
p2 = p2+p2+2p.pcosθ.
p2 = 2p2+2p2cosθ.
-p2 = 2p2cosθ.
cosθ = -1/2.
θ = 120.
p2 = 2p2+2p2cosθ.
-p2 = 2p2cosθ.
cosθ = -1/2.
θ = 120.
Murali Raj K said:
10 years ago
Both the forces are equal thus the resultant is also equal.
Total angle is 360.
Three forces are acting equally thus 360/3=120.
Total angle is 360.
Three forces are acting equally thus 360/3=120.
Krishna said:
1 decade ago
cos(120) = cos(90+30) = -sin30 = -0.5.
Qamar said:
1 decade ago
Let R = f then,
f^2 = f^2+f^2+2f.fcosa.
-f^2/2f^2 = cosa.
cosa = -1/2.
a = 120.
f^2 = f^2+f^2+2f.fcosa.
-f^2/2f^2 = cosa.
cosa = -1/2.
a = 120.
Vishal said:
1 decade ago
Let F1 = 1 and F2 = 1.
We know, cos60 = 1/2.
When 2 forces are acting, both making 60 degrees with the horizontal, we add the angle. (60+60 = 120).
i.e, cos60+cos60 = 1 (which is equal to either of the forces).
We know, cos60 = 1/2.
When 2 forces are acting, both making 60 degrees with the horizontal, we add the angle. (60+60 = 120).
i.e, cos60+cos60 = 1 (which is equal to either of the forces).
Shashi said:
1 decade ago
Can any one can explain in detail?
Himanshu said:
1 decade ago
If the angle would be 120 than only at the mid xcos60+xcos60 = x.
Shereif Wagdi said:
1 decade ago
According to the law of cosines, the formula is:
C^2 = A^2 + B^2 - 2*A*B*Cos θ (notice the negative sign)
The angle is equal to 60 not 120.
That can be easy realized by considering the equilateral triangle, this happens only when all its angles are equal to 60 degrees.
C^2 = A^2 + B^2 - 2*A*B*Cos θ (notice the negative sign)
The angle is equal to 60 not 120.
That can be easy realized by considering the equilateral triangle, this happens only when all its angles are equal to 60 degrees.
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