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 4 of 4.
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.
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°
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.
Musliu said:
10 years ago
The correct answer is 120 bc the magnitude re the same.
So therefore R = P the Law of cosine R^2 = R^2 +R^2 - 2R*R (Cosθ).
R^2 - 2 R^2 = - 2R^2COSθ.
- R^2/R^2 = - 2COSθ.
1/2 = COSθ.
Cos^-10.5 = y.
y = 120.
So therefore R = P the Law of cosine R^2 = R^2 +R^2 - 2R*R (Cosθ).
R^2 - 2 R^2 = - 2R^2COSθ.
- R^2/R^2 = - 2COSθ.
1/2 = COSθ.
Cos^-10.5 = y.
y = 120.
Hamed said:
9 years ago
2 x p x cos 120 = R.
p = R.
p = R.
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