Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Spring constant question?
The bars are in equilibrium, each 2 m long,
and each weighing 73 N. The string pulling
down on the two bars is attached 0.4 m from
the fulcrum on the leftmost bar and 0.3 m
from the left end of the rightmost bar. The
spring (of constant 29 N/cm) is attached at
an angle 39 degrees at the left end of the upper bar.
If the suspended mass is 5 kg, by how much
will the spring stretch? The acceleration of
gravity is 9.8 m/s2 .
Answer in units of cm.
Diagram link: http://tinypic.com/r/18khmu/7
10 points for best answer.
1 Answer
- QuarkLv 610 years agoFavourite answer
From the left to the right on figure :
(2 * 5 * 9.8) * 1.6 + 73 * 0.6 = T1 * 0.4
--> T1 = 501.5 N
T2 sin 39 * 2 = T1 * 1.7 + 73 * 1
--> T2 sin 39 * 2 = 501.5 * 1.7 + 73 * 1
--> T2 = 735.36 N
T2 = F_spring = k * x
735.36 = 2900 * x
--> x= 0.254 m
