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.
Srinu said:
9 years ago
It is based triangle law of forces.
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.
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.
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.
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.
Punith said:
9 years ago
Why 120° why not the answer be 90°?
Mohan said:
8 years ago
Resultant force is equal to either of two forces. Not that all forces have a same magnitude.
Ajay said:
8 years ago
Can anyone tell exact explanation?
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.
Naiya said:
8 years ago
I think we can apply Lami's theorem.
Sachin said:
7 years ago
Based on the Law of parallelogram.
(1)
Ranga... said:
7 years ago
Two force and one resultant.
Than 360/3-120.
Than 360/3-120.
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