Mechanics

https://cdn.discordapp.com/attachments/486844829209198613/513439170983755786/20181117_184107.jpg

Please can someone provide the steps to getting 2Tcostheta in that image? I'm so confused as to how the value was calculated.

Triangles.

Since there are two equal tensions acting on the pulley, the total force on it is $T \cos \theta + T \cos \theta = 2T \cos \theta$ in the direction parallel to the slope.

Ah. I didn't see that, thanks. But I am confused as to why you doubled Tcos(theta). I understand that there are two tension forces but I thought the diagonal line was the resultant force of both them, hence Tcos(theta) wouldn't be doubled.

Ah. I didn't see that, thanks. But I am confused as to why you doubled Tcos(theta). I understand that there are two tension forces but I thought the diagonal line was the resultant force of both them, hence Tcos(theta) wouldn't be doubled.

$T \cos \theta$ is the component of EACH tension along the slope. They ADD to give the resultant along the slope. What's confusing about this?

I am confused because in this similar example:

I have calculated the resultant force like this:

Here I didn't need to double my value.

That triangle is wrong. The tension needs to be hypotenuse. There are also two tensions AGAIN so you need to account for both when considering the resultant force along the slope!

2Tcos45 is the same as T/cos45 though

Yes but the way you laid it out you derived it in the wrong way.

No because if the angle is anything other than 45 degrees your argument fails - you cannot draw on a parallel tension!
Meaning with 45 you just get ‘lucky’ using a faulty approach.

Only the second method is the way to go.

Is this why you don't need to double it?

What I was getting at before is that if you change ONLY one angle to be different from the other, then that approach fails.


But yes, in the case where they are the same we can prove that these approaches are equivalent. Perhaps you have heard of this, but we essentially construct a 'parallelogram' of forces where the slope is the diagonal resultant and we resolve for it.




$F^2 = T^2 + T^2 -2T^2 \cos(180-2\theta) = 2T^2(1 + \cos 2 \theta)$ hence $F = T \sqrt{2}\sqrt{1+\cos 2 \theta}$

Now since $1+\cos 2 \theta = 2\cos^2 \theta$ we get that

$F = F \sqrt{2} [\sqrt{2 \cos^2 \theta}] = 2F \cos \theta$

which is exactly what we have from the usual approach.



Now if the angles are different, then you cannot use this method because the yielded resultant from the parallelogram would not be along the slope. Suppose we slap on the approach above and change one theta to $\alpha$. Then we get

$F^2 = T^2 + T^2 - 2T^2\cos(180-\theta-\alpha) = 2T^2[1+ \cos(\alpha + \theta)]$

But this definitely cannot be rearranged into $T\cos \alpha + T\cos \theta$ from the usual approach.



So it's safe to use the one I was mentioning rather than relying on that one which might trip you up if you don't notice differing angles.
Also your approach doesn't work if we have the same angles, but different 'tensions' on each end which wouldn't happen in these pulley contexts, but just something to look out for in similar situations.