Question

A nonuniform bar is suspended at rest in a horizontal position by two massless cords as shown. One cord makes the angle θ = 36.9° with the vertical; the other makes the angle \(\phi\) = 53.1° with the vertical. If the length L of the bar is 6.10 m, compute the distance x from the left end of the bar to its center of mass.

Answer 1

I'm not sure @Michele_Laino

Answer 2

Net force = 0:\[T_1 \cos \theta + T_2 \cos \phi=mg\]\[T_1 \sin \theta = T_2 \sin \phi\]Net torque about any point is zero. I'm going to take that point as the center of mass.\[T_1 \cos \theta(x) = T_2 \cos \phi (L-x)\]

Answer 3

And 37-53-90 are the angles of the 3-4-5 (or similar derivative) triangles.

Answer 4

so from first equations, i suppose, we have T1 and T2

Answer 5

Yeah, and then we plug and play those values in the equation for torque.

Answer 6

Awesome! let me plugin the numbers... it feels nice to see the textbook answer match with mine

Answer 7

Which textbook is this?

Answer 8

Hey, we don't know mass of the rod ?

Answer 9

Net force = 0:\[T_1 \cos \theta + T_2 \cos \phi=\color{red}{m}g\]\[T_1 \sin \theta = T_2 \sin \phi\]

Answer 10

You're right... this is not as straightforward as I thought. If we try to calculate net torque about other points, maybe it's possible.\[T_2 \cos \phi (L) = mg(x)\]\[T_1 \cos \theta (L) = mg(L-x)\]Again, dunno if anything changes.

Answer 11

halliday
http://as.wiley.com/WileyCDA/WileyTitle/productCd-EHEP002531.html

Answer 12

Ah lol, I have a copy. Never opened the question bank.

Answer 13

it is a good book
at the moment, i am reviewing Chapter 12 Equilibrium and Elasticity

Answer 14

I think we can express both T1 and T2 in terms of m, then plug it in the third eqn

Answer 15

Haha, well, it works.\[mg =\left(\cos \phi + \cot \theta \sin \phi \right) T_2 \]

Answer 16

\[\frac{mg}{T_2} = \frac{L\cos \phi }{x}\]

Answer 17

\[x = \frac{L\cos \phi}{ \cos \phi + \cot \theta \sin \phi }\]

Answer 18

Looks perfect!
http://www.wolframalpha.com/input/?i=solve+u*%5Ccos+%5Ctheta+%2B+v*%5Ccos+%5Cphi%3D10*g%2Cu*%5Csin%5Ctheta+%3D+v*%5Csin%5Cphi%2C+u*x*%5Ccos%5Ctheta+%3D+v*%5Ccos%28%5Cphi%29*%286.1-x%29%2C+g%3D9.8%2C%5Ctheta%3D0.644%2C%5Cphi%3D0.9267