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.
Panneerselvam N said:
4 years ago
To find the resultant
R= √(A2)+(B2)+2ABcosx
As given in the query, the magnitude of the force is the same and let it be f whose resultant be f too.
Substituting the variables in the above equation we get
f= √(f2)+(f2)+2(f2)cosx
Squaring on both sides we get
f2=2(f2)+2(f2)cosx
f2/f2 = 2 + 2 cos x
1 - 2 = 2 cosx
-1/2 = cos x
Therefore x=120°.
R= √(A2)+(B2)+2ABcosx
As given in the query, the magnitude of the force is the same and let it be f whose resultant be f too.
Substituting the variables in the above equation we get
f= √(f2)+(f2)+2(f2)cosx
Squaring on both sides we get
f2=2(f2)+2(f2)cosx
f2/f2 = 2 + 2 cos x
1 - 2 = 2 cosx
-1/2 = cos x
Therefore x=120°.
(12)
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.
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.
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.
S V Manikandan. said:
5 years ago
The two forces acting with the same magnitude and resultant too same magnitude. So the angle between each two forces is equal. So we can divide the circle into three equal segments. The angle of a circle is 360°. Then 360/3 = 120°each.
(2)
Vaibhav said:
5 years ago
The resultant force is the same value of one of this either forces, so we can assume R=P, P1=P, P2=P.
Let using parallelogram therom,
R^2 = P1^2+P2^2+2P1P2COSθ
P^2 = P^2+P^2+2P^2COSθ
P^2 = 2P^2+2P^2COSθ
2P^2COSθ = -P^2
And θ=120°.
Let using parallelogram therom,
R^2 = P1^2+P2^2+2P1P2COSθ
P^2 = P^2+P^2+2P^2COSθ
P^2 = 2P^2+2P^2COSθ
2P^2COSθ = -P^2
And θ=120°.
(3)
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.
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.
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).
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°
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