- #1

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- Homework Statement
- The ABC telescopic extension arm is used to lift a worker to the height of overhead, telephone and electrical cables. For the indicated extension, the centre of gravity of the ##1400\, \textrm{lb}##-arm is at the ##G##-point. The worker, the basket and the fixed equipment in the basket weigh ##450\, \textrm{lb}## and their combined centre of gravity is at point ##C##. For the given position, determine the force exerted at ##B## by the single hydraulic jack using when ##\theta =35\, \textrm{º}##.

Solution: ##\vec{F}_{DB}=6490\, \textrm{lb}\,\,\, 62,1\, \textrm{º}##

- Relevant Equations
- ##\sum M=0##

Figure:

My attempt at a solution:

For the calculation of ##F_{DB}## we consider the equilibrium in the whole machine.

$$\left. \begin{array}{r}

h\rightarrow h=6\cdot \sin \theta +3=6,44\, \textrm{ft} \\

d\rightarrow d=6\cos \theta -1,5=3,41\, \textrm{ft}

\end{array}\right\} \,\, \alpha =\arctan \left( \dfrac{h}{d}\right) =62,1\, \textrm{º}$$

$$\sum M_A=0\rightarrow -9\cos \theta G-20\cos \theta C+F_{DB}\cos \alpha \cdot 3+F_{DB}\sin \alpha \cdot 1,6$$

$$\rightarrow \boxed{F_{DB}=6482,5\, \textrm{lb}}$$

This one is well done, isn't it? It doesn't give me the exact toto but I guess it will be fine.

My attempt at a solution:

For the calculation of ##F_{DB}## we consider the equilibrium in the whole machine.

$$\left. \begin{array}{r}

h\rightarrow h=6\cdot \sin \theta +3=6,44\, \textrm{ft} \\

d\rightarrow d=6\cos \theta -1,5=3,41\, \textrm{ft}

\end{array}\right\} \,\, \alpha =\arctan \left( \dfrac{h}{d}\right) =62,1\, \textrm{º}$$

$$\sum M_A=0\rightarrow -9\cos \theta G-20\cos \theta C+F_{DB}\cos \alpha \cdot 3+F_{DB}\sin \alpha \cdot 1,6$$

$$\rightarrow \boxed{F_{DB}=6482,5\, \textrm{lb}}$$

This one is well done, isn't it? It doesn't give me the exact toto but I guess it will be fine.