3/106 If the weight of the boom is negligible compared with the applied \( 30-\mathrm{kN} \) load, determine the cable tensions \( T_{1} \) and \( T_{2} \) and the force acting at the ball joint at \( A \). Problem 3/106
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- Point A: (0, 0, 0) - Point B: (4, 0, 3) - Point C: (4, 3, 0) - Point D: (0, 0, 3) Show more…
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For the situation drawn in Fig. $5-12(a)$, find $F_{T 1}, F_{T 2}$, and $F_{T 3}$. The boom is uniform and weighs $800 \mathrm{~N}$. First apply the force condition to point- $A$. The appropriate free-body diagram is shown in Fig. $5-12(b)$. We then have $$ F_{T 2} \cos 50.0^{\circ}-2000 \mathrm{~N}=0 \quad \text { and } \quad F_{T 1}-F_{T 2} \sin 50.0^{\circ}=0 $$ From the first of these we find $F_{T 2}=3.11 \mathrm{kN} ;$ then the second equation gives $F_{T 1}=2.38 \mathrm{kN}$. Let us now isolate the boom and apply the equilibrium conditions to it. The appropriate free-body diagram is found in Fig. $5-12(c)$. The torque equation, for torques taken about point $C$, is $$ \text { \& } \sum \tau_{C}=+(L)\left(F_{T 3}\right)\left(\sin 20.0^{\circ}\right)-(L)(3110 \mathrm{~N})\left(\sin 90.0^{\circ}\right)-(L / 2)(800 \mathrm{~N})\left(\sin 40.0^{\circ}\right)=0 $$ Solving for $F_{T 3}$, we compute it to be $9.84 \mathrm{kN}$. If it were required, we could find $F_{R H}$ and $F_{R V}$ by using the $x$ - and $y$ -force equations.
Supratim P.
The 40 -ft boom $A B$ weighs 2 kips; the distance from the axle $A$ to the center of gravity $G$ of the boom is $20 \mathrm{ft}$. For the position shown, determine $(a)$ the tension $T$ in the cable, $(b)$ the reaction at $A .$
Equilibrium of Rigid Bodies
Equilibrium in Two Dimensions
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