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.
PRADEEP KUMAR VERMA said:
1 decade ago
Let forces are which maggnitude 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*PCOSθ
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*PCOSθ
P^2=P^2+P^2+2P^2COSθ
0=1+2COSθ
COSθ=-1/2=COS120.
SO θ=120.
Krushna said:
1 decade ago
Total space angle is 360.
To forces having same magnitude.
And resultant also having same of both.
So 360/3 = 120.
To forces having same magnitude.
And resultant also having same of both.
So 360/3 = 120.
Gunasekaran said:
1 decade ago
Why 360 is divide by 3?
Krunal said:
1 decade ago
Why 360/3?
Sathish said:
1 decade ago
Because two forces and one resultant.
Resultant between two forces.
Resultant between two forces.
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.
Himanshu said:
1 decade ago
If the angle would be 120 than only at the mid xcos60+xcos60 = x.
Shashi said:
1 decade ago
Can any one can explain in detail?
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).
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.
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