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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 agoFavorite 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
