00:01
All right, so let's say we have a package hanging from two supports like this, and this angle phi is different from this angle theta.
00:13
We're told phi is 48 .8 degrees.
00:17
The package has a mass.
00:19
We'll just call it m.
00:21
We want to know what value of theta is the tension in chord two minimized.
00:27
So let's say this is chord one, this is two.
00:29
So the horizontal components, we'll have t1 times the cosine of phi is equal to t2 times the cosine of theta.
00:40
And then we'll also have like t1 times the sign of phi plus t2 times the sine of theta is equal to mg.
00:52
And so if we use the substitution so we can write all this in terms of t2, t2 is just going to be, you know, t1 cosine of phi over the cosine of theta.
01:06
So we'll have t1, sorry, we want to write everything in terms of t2, not t1, so i've got it backwards.
01:18
But if we make the substitution that t1 is t2 times the cosine theta over the cosine of phi, what we'll get is the first term becomes t2.
01:33
Times the tangent of phi times the cosine of theta plus the sign of theta and this is equal to m g so then t2 is equal to basically m g over the tangent of theta or phi times the cosine of theta plus the sign of theta so if we take the derivative of t2 with respect to theta what we'll get is like mg over the tangent of phi times the cosine of theta plus the sign of theta squared with a negative 1 times the derivative of the inside, which is just going to be tangent of phi times the sine of theta, the negative sign plus the cosine of theta...