Chapter 6.3, Problem 6.103P

Knowing that P = 15 lb and Q = 65 lb, determine the components of the forces exerted (a) on member BCDF at C and D, (b) on member ACEG at E.

To determine

The components of the forces exerted on member BCDF at C and D.

Answer to Problem 6.103P

The x component of force at D, $D_x$ is $\underset{\_}{100\text{lb}}$, y component of force at D, $D_y$ is $\underset{\_}{20\text{lb}}$, x component of force at C, $C_x$ is $\underset{\_}{100\text{lb}}$, y component of force at C , $C_y$ is $\underset{\_}{100\text{lb}}$, x component of force at E, $E_x$ is $\underset{\_}{100\text{lb}}$, and  the y component of force at E, $E_y$ is $\underset{\_}{180\text{lb}}$.

Explanation of Solution

Figure 1 is the free body diagram of the entire frame of the system.

Write the equation to find the sum of the moments about point E.

$\sum M_E=\left(P+Q\right)\left(25\text{in}\right)-\left(P+Q\right)\left(15\text{in}\right)-D_x\left(8in\right)=0$ (I)

Here, $P$ is the force acting on point A and F, $Q$ is the force acting on point F and G, $D_x$ is the x component of force at D.

Write the equation to find the sum of x components of force.

$\sum F_x=-E_x+D_x=0$ (II)

Here, $E_x$ is the x component of force at E, and $D_x$ is the x component of force at D.

Consider the member BCDF of the system.

Figure 2 shown below is the free body diagram of member BCDF.

Write the equation to find the sum of x component of force.

$\sum F_x=-C_x+D_x=0$ (III)

Here, $C_x$ is the x component of force at C, and $D_x$ is the x component of force at point D.

Write the equation to find the sum of moments at D.

$\sum M_D=-E_x\left(4\text{in}\right)-P\left(15\text{in}\right)+Q\left(25\text{in}\right)+C_y\left(10\text{in}\right)=0$ (IV)

Here, $E_x$ is the x component of force at E, $P$ is the force at P, $Q$ is the force at Q, and $C_y$ is the y component of force at C.

Write the equation to find the sum of y components of force.

$\sum F_y=D_y+C_y=0$ (V)

Here, $D_y$ is the y component of force at D, and $C_y$ is the y component of force at C.

Consider the free body diagram of member ACEG. The free body diagram of ACEG is given in figure 3.

Write the equation to find the sum of y component of force.

$\sum F_y=E_y-C_y-P-Q=0$ (VI)

Here, $E_y$ is the y component of force at E, $C_y$ is the y component of force at C, $P$ is the force at P, and $Q$ is the force at Q.

Conclusion:

Solve equation (I) to get the value of $D_x$.

$D_x=\left(P+Q\right)\frac{10}{8}$

Substitute $15\text{lb}$ for $P$, and $65\text{lb}$ for $Q$ to get $D_x$.

$\begin{array}{c}{D}_x=\left(15+65\right)\frac{10}{8}\\ =100\text{lb}\end{array}$

Solve equation (III) to get the value of $C_x$.

$\begin{array}{c}{C}_x=D_x\\ =\left(P+Q\right)\frac{10}{8}\end{array}$

Substitute $15\text{lb}$ for $P$, and $65\text{lb}$ for $Q$ to get $C_x$.

$\begin{array}{c}{C}_x=\left(15+65\right)\frac{10}{8}\\ =100\text{lb}\end{array}$

Substitute $\left(P+Q\right)\frac{10}{8}$ for $D_x$ in equation (IV) to get $C_y$.

Substitute $15\text{lb}$ for $P$, and $65\text{lb}$ for $Q$ to get $C_y$.

$\begin{array}{c}{C}_y=2\left(65\right)-2\left(15\right)\\ =100\text{lb}\end{array}$

Substitute $2Q-2P$ for $C_y$ in equation (V) to get $D_y$.

$\begin{array}{c}{D}_y+\left(2Q-2P\right)=0\\ P-Q=0\\ D_y=-Q+3P\end{array}$

Substitute $15\text{lb}$ for $P$, and $65\text{lb}$ for $Q$ to get $D_y$.

$\begin{array}{c}{D}_y=-65+3\left(15\right)\\ =-20\text{lb}\end{array}$

Therefore, The x component of force at D, $D_x$ is $\underset{\_}{100\text{lb}}$, y component of force at D, $D_y$ is $\underset{\_}{20\text{lb}}$, x component of force at C, $C_x$ is $\underset{\_}{100\text{lb}}$, y component of force at C , $C_y$ is $\underset{\_}{100\text{lb}}$,

To determine

The component of force on member ACEG at E.

Answer to Problem 6.103P

The x component of force at E, $E_x$ is $\underset{\_}{100\text{lb}}$, and  the y component of force at E, $E_y$ is $\underset{\_}{180\text{lb}}$.

Explanation of Solution

Rewrite equation (II) to find the x component of force at E.

$\sum F_x=-E_x+D_x=0$ (VII)

Here, $E_x$ is the x component of force at E, and $D_x$ is the x component of force at D.

Rewrite equation (VI) to get the y component of force at E.

$\sum F_y=E_y-C_y-P-Q=0$ (VIII)

Here, $E_y$ is the y component of force at E, $C_y$ is the y component of force at C, $P$ is the force at P, and $Q$ is the force at Q.

Conclusion:

Substitute $\left(P+Q\right)\frac{10}{8}$ for $D_x$ in equation (VII) to get $E_x$.

$\begin{array}{l}-E_x\left(P+Q\right)\frac{10}{8}=0\\ E_x=\left(P+Q\right)\frac{10}{8}\end{array}$

Substitute $15\text{lb}$ for $P$, and $65\text{lb}$ for $Q$ to get $E_x$.

$\begin{array}{c}{E}_x=\left(15+65\right)\frac{10}{8}\\ =100\text{lb}\end{array}$

Substitute $2Q-P$ for $C_y$ in equation (VIII) to get $E_y$.

$\begin{array}{c}{E}_y-\left(2Q-2P\right)-P-Q=0\\ E_y=3Q-P\end{array}$

Substitute $15\text{lb}$ for $P$, and $65\text{lb}$ for $Q$ to get $E_y$.

$\begin{array}{c}{E}_y=3\left(65\right)-15\\ =180\text{lb}\end{array}$

Therefore, The x component of force at E, $E_x$ is $\underset{\_}{100\text{lb}}$, and  the y component of force at E, $E_y$ is $\underset{\_}{180\text{lb}}$.