The bars are in equilibrium, each 2 m long, and each weighing 92 N. The string p

Question

The bars are in equilibrium, each 2 m long, and each weighing 92 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 27 N/cm) is attached at
an angle 29 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.

Please provide a step-by-step solution using equations with assigned variables to make it easy to follow. Thanks in advance!

The bars are in equilibrium, each 2 m long, and each weighing 92 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 27 N/cm) is attached at an angle 29 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.

Explanation / Answer

now one thing to rememebr is that there is no movement in this setup, its all at equilibrium, and the value of acceleration (acceleration that is visible, not gravitational) of any component is 0,
let bar 1 be the one in the middle, and bar 2 with the spring and attached to the wall
take the 5 kg weight

let there be a tension T1 in that rope
5g = T1
force on pulley is 2 T1

force on bar is 2 T1 = 10g

moment around the fulcrum
M1 = 10g * (2-0.4) = 0.4*T2
T2 = 40g

now this creates a moment M2 around fulcrum of bar 2
M2 = 40g * (2-0.3) = Fs sin(29) * 2 .... Fs is spring force, and i am only taking theperpendicular component which is the one that causes the moment

Fs = 70.13 N
x = Fs/k = 54.026/27 = 2.597 cm ... which is extension in the spring

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