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Question 6.123: The spring has an unstretched length of 0.3 m. Determine the...

The spring has an unstretched length of 0.3 m. Determine the mass m of each uniform link if the angle θ = 20° for equilibrium.

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y2(0.6)=sin⁡20∘ \frac{y}{2(0.6)}=\sin 20^{\circ}

 

y=1.2sin⁡20∘y=1.2 \sin 20^{\circ}

 

Fs=(1.2sin⁡20∘−0.3)(400)=44.1697 N F_s=\left(1.2 \sin 20^{\circ}-0.3\right)(400)=44.1697 \mathrm{~N}

 

⤹+ΣMA=0;Ex(1.4sin⁡20∘)−2(mg)(0.35cos⁡20∘)=0 ⤹ +\Sigma M_A=0 ; \quad E_x\left(1.4 \sin 20^{\circ}\right)-2(m g)\left(0.35 \cos 20^{\circ}\right)=0

Ex=1.37374(mg)E_x=1.37374(\mathrm{mg})

⤹+ΣMC=0;1.37374mg(0.7sin⁡20∘)+mg(0.35cos⁡20∘)−44.1697(0.6cos⁡20∘)=0⤹ +\Sigma M_C=0 ; \quad 1.37374 m g\left(0.7 \sin 20^{\circ}\right)+m g\left(0.35 \cos 20^{\circ}\right)-44.1697\left(0.6 \cos 20^{\circ}\right)=0

mg=37.860 m g=37.860

m=37.860/9.81=3.86 kgm=37.860 / 9.81=3.86 \mathrm{~kg}

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