Thread: 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

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 ft

stretch, ft

obtuse angle at

using net torque about point A ...



lbs/ft

Originally Posted by skeeter

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



stretched spring length ft

stretch, ft

obtuse angle at

using net torque about point A ...



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?

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 , and the 3 ft rod ... the fundamental setup for the law of cosines.