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
30°
60°
90°
120°
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
35 comments Page 2 of 4.

Mohan said:   8 years ago
Resultant force is equal to either of two forces. Not that all forces have a same magnitude.

Punith said:   9 years ago
Why 120° why not the answer be 90°?

Jitendra mittal said:   9 years ago
Let forces are which magnitude P.

SO, GIVEN IN Q. RESULTANT = P.
SO WE KNOW THAT RESULTANT.

R^2 = A^2+B^2+2ABCOSθ THEN
P^2 = P^2+P^2+2P^2COSθ.
P^2 = P^2+P^2+2P^2COSθ.

0=1+2COSθ.

COSθ=-1/2=COS120.

SO, θ=120.

Komal said:   9 years ago
It is based on the law of parallelogram.

Two adjacent sides having forces say 'P' & 'Q' then resultant 'R' is given by,

R^2 = P^2 + Q^2 + 2PQ Cosθ.

Solving this equation taking P = Q will get answer as 120.

Srinu said:   9 years ago
It is based triangle law of forces.

Dhanendra said:   9 years ago
Correct answer will be 60θ as in fcosθ plus fcosθ equal to f which gives θ equal to 60θ.

P.Chilambarasan said:   9 years ago
R^2 = P^2 + q^2 + 2pq Cosθ.
R = P = Q = F.

Substitute F in the equation, you get Cosθ = -1.
So, θ = 180°.

Ankit ughade said:   9 years ago
For getting the equilibrium condition all these forces must have the same angle to each other.

Hence 360/3 = 120.

Hamed said:   9 years ago
2 x p x cos 120 = R.
p = R.

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.


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