# Thread: another simple statics problem

1. ## another simple statics problem

Unfortunately, there are no examples in the book like this and I don't have any physics books.

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

2. Originally Posted by hairy
Unfortunately, there are no examples in the book like this and I don't have any physics books.

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

3. 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?

4. 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.