Finding the work done by Spiderman

[jbriggs444]

Is my proof for θ=90+ϕ geometrically correct?

Yes. The concern with differentials is distracting and the drawings could be done more clearly -- in particular, labelling the angle you are calling $\theta$. But I agree with the conclusion.

It might make it easier to evaluate the resulting integral if you would write $\cos ( 90 + \phi )$ as $-\sin \phi$.

Personally, I would have looked at the formula for the projection of a vector normal to a [oriented] surface. That involves $\sin \theta$ where $\theta$ is the angle the vector makes with respect to the surface. In this case, the displacement vector makes an angle of $- \phi$ with respect to a horizontal surface. The projection thus involves $\sin -\phi = - \sin \phi$.

You can get the dot product of two vectors by taking the projection of one in the direction of the other and multiplying the magnitude of the one by the [signed] magnitude of the projection of the other.

Actually, I would have been sloppy with the sign convention, ignored the orientation of the surface, looked at $\sin \phi$ and then reasoned that the displacement and gravity are pointing generally away from each other and inverted the sign of the resulting path integral.

Actually, I would have recognized a conservative field, reasoned that all path integrals between a particular pair of endpoints across a conservative field will yield the same result and either picked a trivial path or computed the potential difference.

But we can definitely get to work evaluating the integral.

 

[member 731016]

Yes. The concern with differentials is distracting and the drawings could be done more clearly -- in particular, labelling the angle you are calling $\theta$. But I agree with the conclusion.

It might make it easier to evaluate the resulting integral if you would write $\cos ( 90 + \phi )$ as $-\sin \phi$.

Personally, I would have looked at the formula for the projection of a vector normal to a [oriented] surface. That involves $\sin \theta$ where $\theta$ is the angle the vector makes with respect to the surface. In this case, the displacement vector makes an angle of $- \phi$ with respect to a horizontal surface. The projection thus involves $\sin -\phi = - \sin \phi$.

You can get the dot product of two vectors by taking the projection of one in the direction of the other and multiplying the magnitude of the one by the [signed] magnitude of the projection of the other.

Actually, I would have been sloppy with the sign convention, ignored the orientation of the surface, looked at $\sin \phi$ and then reasoned that the displacement and gravity are pointing generally away from each other and inverted the sign of the resulting path integral.

Actually, I would have recognized a conservative field, reasoned that all path integrals between a particular pair of endpoints across a conservative field will yield the same result and either picked a trivial path or computed the potential difference.

But we can definitely get to work evaluating the integral.

Thank you so much for your reply @jbriggs444 !

Yeah true, there is a much easier method if I had recognized the conservative field! Sorry, what did you find distracting about my concern with differentials?

Was it the fact that first method I had to make an approximation and the second I did not (due to the choice of triangles)

I will post the integral

Many thanks!

 

[member 731016]

Here is the integration @jbriggs444 !

$W = -Fr\int_0^{60} \sin\phi d\phi$
$W = mgr\cos60= \frac {mgr}{2}$ which is correct!

Many thanks!

 

[jbriggs444]

Yeah true, there is a much easier method if I had recognized the conservative field! Sorry, what did you find distracting about my concern with differentials?

Well, a drawing of an isosceles triangle with two right angles is a bit distracting.

 

[member 731016]

Well, a drawing of an isosceles triangle with two right angles is a bit distracting.

Oh true! Thank for letting me know @jbriggs444 !