# another simple statics problem

• October 16th 2009, 04:33 PM
hairy
another simple statics problem
Unfortunately, there are no examples in the book like this and I don't have any physics books.
http://dl.getdropbox.com/u/594924/5-54.jpg

The answer in the back of the book says k = 11.2 lb/ft. But the spring is only stretched an additional .464 feet when theta is 30, so is that 5.104 pounds all going to the y component to hold up the rod? What is the method here, and how many "steps" does it take?

Thanks
• October 16th 2009, 07:09 PM
skeeter
Quote:

Originally Posted by hairy
Unfortunately, there are no examples in the book like this and I don't have any physics books.
http://dl.getdropbox.com/u/594924/5-54.jpg

The answer in the back of the book says k = 11.2 lb/ft. But the spring is only stretched an additional .464 feet when theta is 30, so is that 5.104 pounds all going to the y component to hold up the rod? What is the method here, and how many "steps" does it take?

Thanks

seems the geometry is a bit more complex than first meets the eye ...

stretched spring length $L = \sqrt{6^2+3^2 - 2(6)(3)\cos(30)} \approx 3.718$ ft

stretch, $x \approx 0.718$ ft

obtuse angle at $B = 180 - \arcsin\left(\frac{3}{L}\right) \approx 126.2^\circ$

using net torque about point A ...

$kx(3)\sin(B) = 15(1.5)\sin(60)$

$k = \frac{5(1.5)\sin(60)}{x\sin(B)} \approx 11.2$ lbs/ft
• October 17th 2009, 12:40 AM
hairy
Quote:

Originally Posted by skeeter
seems the geometry is a bit more complex than first meets the eye ...

stretched spring length $L = \sqrt{6^2+3^2 - 2(6)(3)\cos(30)} \approx 3.718$ ft

stretch, $x \approx 0.718$ ft

obtuse angle at $B = 180 - \arcsin\left(\frac{3}{L}\right) \approx 126.2^\circ$

using net torque about point A ...

$kx(3)\sin(B) = 15(1.5)\sin(60)$

$k = \frac{5(1.5)\sin(60)}{x\sin(B)} \approx 11.2$ lbs/ft

thanks a lot. i never would've gotten this. the law of cosines is not one of the first things i check out.

edit: would you mind explaining what about this problem indicated to you that this was the strategy?
• October 17th 2009, 12:55 AM
skeeter
Quote:

Originally Posted by hairy
thanks a lot. i never would've gotten this. the law of cosines is not one of the first things i check out.

edit: would you mind explaining what about this problem indicated to you that this was the strategy?

I just saw the Side-Angle-Side setup with the 6 ft horizontal spacing, angle $\theta$ , and the 3 ft rod ... the fundamental setup for the law of cosines.