Chapter 10, Problem 10.101RP

Determine the vertical force P that must be applied at G to maintain the equilibrium of the linkage.

Fig. P10.101 and P10.102

To determine

Find the vertical force P that should be applied at G for which the equilibrium is maintained.

Answer to Problem 10.101RP

The vertical force P required for the equilibrium is $\underset{\_}{60.0\text{lb}(\downarrow )}$.

Explanation of Solution

Show the free-body diagram of the arrangement as in Figure 1.

Consider the vertical displacement at A as $\left(\delta y_A\downarrow \right)$.

Find the vertical displacement $\left(\delta y_C\right)$ at C as follows;

$\begin{array}{l}\frac{8}{\delta y_A}=\frac{12}{\delta y_C}\\ \delta y_C=\frac{12}{8}\delta y_A\\ \delta y_C=1.5\delta y_A(\uparrow )\end{array}$

Find the vertical displacement $\left(\delta y_E\right)$ at E as follows;

$\delta y_E=\delta y_C=1.5\delta y_A(\uparrow )$

Find the vertical displacement $\left(\delta y_D\right)$ at D as follows;

$\begin{array}{c}\frac{18}{\delta y_D}=\frac{6}{\delta y_E}\\ \delta y_D=\frac{18}{6}\delta y_E\\ \delta y_D=3\left(1.5\delta y_A\right)\\ =4.5\delta y_A(\uparrow )\end{array}$

Find the vertical displacement $\left(\delta y_G\right)$ at G as follows;

$\begin{array}{c}\frac{18}{\delta y_D}=\frac{10}{\delta y_G}\\ \delta y_G=\frac{10}{18}\delta y_D\\ \delta y_G=\frac{10}{18}\left(4.5\delta y_A\right)\\ =2.5\delta y_A(\downarrow )\end{array}$

Use the virtual work principle:

$\begin{array}{l}\delta U=0\\ 300\delta y_A-100\delta y_D+P\delta y_G=0\end{array}$

Substitute $4.5\delta y_A$ for $\delta y_D$ and $2.5\delta y_A$ for $\delta y_G$.

$\begin{array}{l}300\delta y_A-100\left(4.5\delta y_A\right)+P\left(2.5\delta y_A\right)=0\\ 300-450+2.5P=0\\ 2.5P=150\\ P=60.0\text{lb}(\downarrow )\end{array}$

Therefore, the vertical force P required for the equilibrium is $\underset{\_}{60.0\text{lb}(\downarrow )}$.