Question 6.SP.7: A hydraulic-lift table is used to raise a 1000-kg crate. The......
A hydraulic-lift table is used to raise a 1000-kg crate. The table consists of a platform and two identical linkages on which hydraulic cylinders exert equal forces. (Only one linkage and one cylinder are shown.) Members EDB and CG are each of length 2a, and member AD is pinned to the midpoint of EDB. If the crate is placed on the table so that half of its weight is supported by the system shown, determine the force exerted by each cylinder in raising the crate for θ = 60°, a = 0.70 m, and L = 3.20 m. Show that the result is independent of the distance d.
STRATEGY: The free-body diagram of the entire frame will involve more than three unknowns, so it alone can not be used to solve this problem. Instead, draw free-body diagrams of each component of the machine and work from them.
MODELING: The machine consists of the platform and the linkage. Its free-body diagram (Fig. 1) includes an input force F_{DH} exerted by the cylinder; the weight W/2, which is equal and opposite to the output force; and reactions at E and G, which are assumed to be directed as shown. Dismember the mechanism and draw a free-body diagram for each of its component parts (Fig. 2). Note that AD, BC, and CG are two-force members. Member CG has already been assumed to be in compression; now assume that AD and BC are in tension and direct the forces exerted on them as shown. Use equal and opposite vectors to represent the forces exerted by the two-force members on the platform, on member BDE, and on roller C.
ANALYSIS:
Free Body: Platform ABC (Fig. 3).
\underrightarrow{+} \begin{array}{c c}{{\Sigma F_{x}=0\colon\qquad}}&{{\qquad F_{A D}\cos\theta=0}}\end{array}\qquad F_{A D}=0+↑\begin{array}{c l l l}{\Sigma F_{y}=0\colon\qquad B+C-{\frac{1}{2}}W=0\qquad B+C={\frac{1}{2}}W}\end{array}\qquad (1)
Free Body: Roller C (Fig. 4). Draw a force triangle and obtain F_{BC} = C cot θ.
Free Body: Member BDE (Fig. 5). Recalling that F_{AD}, = 0, you have
+↺\Sigma M_{E}=0\colon~~F_{D H}\cos(\phi-90^{\circ})a-B(2a\cos\theta)-F_{B C}(2a\sin\theta)=0
F_{D H}a\,\sin\phi-B(2a\cos\theta)-(C\cot\theta)(2a\sin\theta)=0
F_{D H}\sin\phi-2(B+C)c o s\theta=0
From Eq. (1), you obtain
F_{D H}=W {\frac{\mathrm{cos~\theta}}{\mathrm{sin~\phi}}} (2)
Note that the result obtained is independent of d. ◂
Applying first the law of sines to triangle EDH (Fig. 6), you have
{\frac{\sin\phi}{E H}}={\frac{\sin\theta}{D H}}\;\;\;\sin\phi={\frac{E H}{D H}}\sin\theta (3)
Now using the law of cosines, you get
(DH )² = a² + L² − 2aL cos θ
= ( 0.70 )² + (3.20 )² − 2 ( 0.70 )( 3.20 ) cos 60°
(DH )² = 8.49 DH = 2.91 m
Also note that
W = mg = (1000 kg)(9.81 m/s²) = 9810 N = 9.81 kN
Substituting for sin ϕ from Eq. (3) into Eq. (2) and using the numerical data, your result is
F_{D H}=W\frac{D H}{E H}\mathrm{cot}\theta=(9.81~\mathrm{kN})\frac{2.91~\mathrm{m}}{3.20~\mathrm{m}}\mathrm{cot}\ 60^{\circ}
F_{D H} = 5.15 kN ◂
REFLECT and THINK: Note that link AD ends up having zero force in this situation. However, this member still serves an important function, because it is necessary to enable the machine to support any horizontal load that might be exerted on the platform.
