The members of a truss are pin connected at joint O . Determine the magnitudes of F1 and F2 for equilibrium. Set θ=60.
Figure 3.1
Solution:
Free-body diagram:
Equations of Equilibrium:
Take the sum of horizontal forces considering forces to the right positive, and equate to zero.
∑ F x = 0 F 1 cos 60 ° + F 2 sin 70 ° − 5 cos 30 ° − 4 5 ( 7 ) = 0 0.5 F 1 + 0.9397 F 2 = 9.9301 ( 1 ) \begin {aligned}
\sum{F}_x &= 0 & \\
F_1 \cos{60 \degree}+F_2 \sin{70 \degree}-5\cos{30 \degree}-\dfrac{4}{5}\left(7\right) &= 0 &\\
0.5F_1+0.9397F_2&=9.9301 &(1)\\
\end {aligned}
∑ F x F 1 cos 60° + F 2 sin 70° − 5 cos 30° − 5 4 ( 7 ) 0.5 F 1 + 0.9397 F 2 = 0 = 0 = 9.9301 ( 1 )
Take the sum of vertical forces considering upward forces positive, and equate to zero.
∑ F y = 0 − F 1 sin 60 ° + F 2 cos 70 ° + 5 sin 30 ° − 3 5 ( 7 ) = 0 − 0.8660 F 1 + 0.3420 F 2 = 1.7 ( 2 ) \begin{aligned}
\sum F_y&=0 &\\
-F_1\sin60\degree+F_2\cos70\degree+5\sin30\degree-\dfrac{3}{5}\left(7\right)&=0 &\\
-0.8660F_1+0.3420F_2&=1.7 &(2)\\
\end{aligned} ∑ F y − F 1 sin 60° + F 2 cos 70° + 5 sin 30° − 5 3 ( 7 ) − 0.8660 F 1 + 0.3420 F 2 = 0 = 0 = 1.7 ( 2 )
Now, we have two equations with two unknowns F 1 F_1 F 1 and F 2 F_2 F 2 . So, we have a system of two equations. We can solve this using algebra, or we can directly use our calculator with this capability. The answers are
F 1 = 1.83 kN F 2 = 9.60 kN F_1=1.83 \: \text{kN}\\
F_2=9.60 \: \text{kN} F 1 = 1.83 kN F 2 = 9.60 kN