- #1
Hernaner28
- 263
- 0
Hi. I have a concrete doubt with this problem. Here's the pic.
It asks me to calculate the work done by force P (the ball moves with constant speed). So the solution is in the book and I understood everything, but the problem comes here,
the force P in axis y is zero so the work of P should be:
[tex]\int\limits_{{x_o}}^{{x_f}} {{P_x}dx} [/tex]
And we know that Px is equal to the tension in axis x so:
[tex]\int\limits_{}^{} {T\sin \theta dx} [/tex]
But we need to convert the variable dx into theta variable. And the books states that as:
[tex]x = L\sin \theta [/tex]
then:
[tex]dx = L\cos \theta d\theta [/tex]
But shouldn't it be:
[tex]dx = L\sin\theta d\theta [/tex]??
Why did it take the derivative of sine and not of x on the other side? Thank you!
It asks me to calculate the work done by force P (the ball moves with constant speed). So the solution is in the book and I understood everything, but the problem comes here,
the force P in axis y is zero so the work of P should be:
[tex]\int\limits_{{x_o}}^{{x_f}} {{P_x}dx} [/tex]
And we know that Px is equal to the tension in axis x so:
[tex]\int\limits_{}^{} {T\sin \theta dx} [/tex]
But we need to convert the variable dx into theta variable. And the books states that as:
[tex]x = L\sin \theta [/tex]
then:
[tex]dx = L\cos \theta d\theta [/tex]
But shouldn't it be:
[tex]dx = L\sin\theta d\theta [/tex]??
Why did it take the derivative of sine and not of x on the other side? Thank you!