Question

A block (m=25 kg) is held in equilibrium by two springs as shown in the figure. Calculate in this condition the unstretched length of the springs.

Data h₁ = 2,5, m, k₁ = 250 N/m, h₂ = 1,5 m, k₂ = 300 N/m

A block (m-25 kg) is held in equilibrium by two springs as shown in the figure. Calculate in this condition the unstretched length of the springs. Data: hi = 2,5 m, ki = 250 N/m, h, = 1,5 m, k,-300 N/m. k2 h2 55% 65°

Answer 1

lIX fi1 1 65 1x F2x mg

Now From triagle OAB length of stretched spring 1

$L_{1}= OB = \frac{AB}{\sin 55^{0}}= \frac{2.5 m}{\sin 55^{0}}= 3.052 m$

Now From triagle OCD length of stretched spring 2

$L_{2}= OD = \frac{CD}{\sin 65^{0}}= \frac{1.5 m}{\sin 65^{0}}= 1.655 m$

Now From the FBD, Using translational equilibrium along X as well as Y axis

$k_{1} \Delta L_{1}\cos 55^{0}= k_{2}\Delta L_{2}\cos 65^{0}$

$250 N/m \Delta L_{1}\cos 55^{0}= 300 N/m\Delta L_{2}\cos 65^{0}$

$\Delta L_{1}= \frac{300 N/m\Delta L_{2}\cos 65^{0}}{250 N/m\cos 55^{0}}= 0.8842\Delta L_{2}$

$k_{1}\Delta L_{1}\sin 55^{0}+k_{2}\Delta L_{2}\sin 65^{0}= mg$

$250 N/m\Delta L_{1}\sin 55^{0}+300 N/m\Delta L_{2}\sin 65^{0}= 25kg\times 9.8m/s^{2}$

$204.8\Delta L_{1}+271.9\Delta L_{2}= 245$

using value from equation above

$204.8\times 0.8842\Delta L_{2}+271.9\Delta L_{2}= 245$

$\Delta L_{2}= \frac{245}{204.8\times 0.8842+271.9}= 0.5409 m$

and

$\Delta L_{1}= 0.8842\times 0.5409 m = 0.4782 m$

Finally

length of unstretched spring 1

$L_{1}' = L_{1}-\Delta L_{1}= 3.052 m-0.4782 m= 2.574 m$

length of unstretched spring 2

$L_{2}'= L_{2}-\Delta L_{2}= 1.655 m-0.5409 m= 1.1141 m$