Question #79986
Two equal forces have the square of their resultant equal to three times their product. Find the angle between them.
Expert's answer
The resultant of two equal forces
R=F_1+F_2
Let the angle between forces is θ. Then
R^2=(F_1+F_2 )^2=F_1^2+F_2^2+2F_1 F_2 cos〖θ=2F^2+2F^2 cosθ 〗
The product of two equal forces is equal to F^2
Since
R^2=3F^2
we get
2F^2+2F^2 cosθ=3F^2
Finally
cosθ=1/2
θ=60°
R=F_1+F_2
Let the angle between forces is θ. Then
R^2=(F_1+F_2 )^2=F_1^2+F_2^2+2F_1 F_2 cos〖θ=2F^2+2F^2 cosθ 〗
The product of two equal forces is equal to F^2
Since
R^2=3F^2
we get
2F^2+2F^2 cosθ=3F^2
Finally
cosθ=1/2
θ=60°
