- #1
RK1992
- 89
- 0
Given:
[tex]T=mgcos\theta[/tex]
and
[tex] cos^2\theta + \frac{mV^2}{ag} cos\theta - 1 = 0 [/tex]
show that:
[tex] T = \frac{mV^2}{2a} [1+\sqrt{1+(\frac{2ag}{V^2})^2}] [/tex]
I always get:
[tex] T = - \frac{mV^2}{2a} [1 + \sqrt{1 + (\frac{2ag}{V^2})^2}] [/tex]
T represents a tension in a string so I understand why you'd take just the positive root (can't have a negative tension in a string) but I don't see how you can get a positive mV²/2a out in the front.. Any ideas if it's me who's wrong or the question? If it is impossible to rearrange so there's not a negative outside, do you just take the magnitude for the reason I stated, because it's a string?
[tex]T=mgcos\theta[/tex]
and
[tex] cos^2\theta + \frac{mV^2}{ag} cos\theta - 1 = 0 [/tex]
show that:
[tex] T = \frac{mV^2}{2a} [1+\sqrt{1+(\frac{2ag}{V^2})^2}] [/tex]
I always get:
[tex] T = - \frac{mV^2}{2a} [1 + \sqrt{1 + (\frac{2ag}{V^2})^2}] [/tex]
T represents a tension in a string so I understand why you'd take just the positive root (can't have a negative tension in a string) but I don't see how you can get a positive mV²/2a out in the front.. Any ideas if it's me who's wrong or the question? If it is impossible to rearrange so there's not a negative outside, do you just take the magnitude for the reason I stated, because it's a string?
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