Chapter 6.3, Problem 6.88P

The 48-lb load can be moved along the line of action shown and applied at A, D, or E. Determine the components of the reactions at B and F if the 48-lb load is applied (a) at A, (b) at D, (c) at E.

Fig. P6.88 and P6.89

SOLUTION

Free body: Entire frame:

The following analysis is valid for (a), (b) and (c) since the position of the load along its line of action is immaterial.

∑M_F = 0: (48 lb)(8 in.) − B_x (12 in.) = 0

B_x = 32 lb B_x = 32 lb →

∑F_x = 0: 32 lb + F_x = 0

F_x = −32 lb F_x = 32 lb ←

+↑ ∑F_y = 0: B_y + F_y  − 48 lb = 0    (1)

a) Load applied at A.

Free body: Member CDB

CDB is a two-force member. Thus, the reaction at B must be directed along BC.

$\frac{B_y}{\text{32 lb}}=\frac{5\text{ in}\text{.}}{16\text{ in}\text{.}}$ B_y = 10 lb ↑

From Eq. (1): 10 lb + F_y − 48 lb = 0

F_y = 38 lb F_y = 38 lb ↑

Thus reactions are:

B_x = 32.0 lb →, B_y = 10.00 lb ↑

F_x = 32.0 lb ←, F_y = 38.00 lb ↑

(b) Load applied at D.

Free body: Member ACF.

ACF is a two-force member. Thus, the reaction at F must be directed along CF.

$\frac{F_y}{\text{32 lb}}=\frac{7\text{ in}\text{.}}{16\text{ in}\text{.}}$ F_y = 14 lb ↑

From Eq. (1): B_y + 14 lb – 48 lb = 0

B_y = 34 lb, B_y = 34 lb ↑

Thus, reactions are:

B_x = 32.0 lb ←, B_y = 34.00 lb ↑

F_x = 32.0 lb →, F_y = 14.00 lb ↑

(c) Load applied at E.

Free body: Member CDB.

This is the same free body as in Part (a).

Reactions are same as (a)

To determine

The components of reaction at $B$ and $F$ if the $48\text{lb}$ load is applied at $A$.

Answer to Problem 6.88P

The components of reaction at $B$ and $F$ if the $48\text{lb}$ load is applied at $A$ are $\underset{\_}{B_x=32.0\text{lb}\to }$, $\underset{\_}{B_y=10.0\text{lb}\uparrow }$, $\underset{\_}{F_x=32.0\text{lb}\leftarrow }$, and $\underset{\_}{F_y=38.0\text{lb}\uparrow }$.

Explanation of Solution

The arrangement is shown in Fig. P6.88. The free-body diagram of the entire frame is given in Figure 1.

Write the expressions for equilibrium for the given system using the conditions on moments $M$ and force $F$.

The counter clockwise moments at $F$ sum up to zero.

$\Sigma M_F=0$ (I)

The sum of $x$ component of all forces is equal to zero.

$\Sigma F_x=0$ (II)

The sum of $y$ component of all forces is equal to zero.

$\Sigma F_y=0$ (III)

The load is applied at $A$. The free-body diagram of the member $CDB$ is given in Figure 2. Since $CDB$ is a two-force member, the reaction at $B$ is directed along $BC$.

Conclusion:

Apply the condition in equation (I) to the free-body diagram in Figure 1 and solve for $B_x$.

$\begin{array}{c}\left(48\text{lb}\right)\left(8\text{in}\text{.}\right)-B_x\left(12\text{in}\text{.}\right)=0\\ B_x=32\text{lb}\\ B_x=32\text{lb}\to \end{array}$

Apply the condition in equation (II) to the free-body diagram in Figure 1 and solve for $F_x$.

$\begin{array}{c}32\text{lb}+F_x=0\\ F_x=-32\text{lb}\\ F_x=32\text{lb}\leftarrow \end{array}$

Apply the condition in equation (III) to the free-body diagram in Figure 1.

$B_y+F_y-48\text{lb}=0$ (IV)

From the free-body diagram in Figure 2, write the expression connecting the components of the reaction $B$ with the horizontal and vertical separations between the points $C$ and $B$.

$\frac{B_x}{16\text{in}\text{.}}=\frac{B_y}{5\text{in}\text{.}}$ (V)

Substitute $32\text{lb}$ for $B_x$ in equation (V) and solve for $B_y$.

$\begin{array}{c}\frac{32\text{lb}}{16\text{in}\text{.}}=\frac{B_y}{5\text{in}\text{.}}\\ B_y=\left(\frac{5\text{in}\text{.}}{16\text{in}\text{.}}\right)32\text{lb}\\ B_y=10\text{lb}\\ B_y=10.0\text{lb}\uparrow \end{array}$

Substitute $10\text{lb}$ for $B_y$ in equation (IV) and solver for $F_y$.

$\begin{array}{c}10\text{lb}+F_y-48\text{lb}=0\\ F_y=38\text{lb}\\ F_y=38.0\text{lb}\uparrow \end{array}$

Therefore, the components of reaction at $B$ and $F$ if the $48\text{lb}$ load is applied at $A$ are $\underset{\_}{B_x=32.0\text{lb}\to }$, $\underset{\_}{B_y=10.0\text{lb}\uparrow }$, $\underset{\_}{F_x=32.0\text{lb}\leftarrow }$, and $\underset{\_}{F_y=38.0\text{lb}\uparrow }$.

To determine

The components of reaction at $B$ and $F$ if the $48\text{lb}$ load is applied at $D$.

Answer to Problem 6.88P

The components of reaction at $B$ and $F$ if the $48\text{lb}$ load is applied at $D$ are $\underset{\_}{B_x=32.0\text{lb}\leftarrow }$, $\underset{\_}{B_y=34.0\text{lb}\uparrow }$, $\underset{\_}{F_x=32.0\text{lb}\to }$, and $\underset{\_}{F_y=14.0\text{lb}\uparrow }$.

Explanation of Solution

The $x$ components of the forces at $B$ and $F$ are obtained from the free-body diagram given in Figure 1.

$B_x=32.0\text{lb}\leftarrow$

$F_x=32.0\text{lb}\to$

The load is applied at $D$. The free-body diagram of the member $ACF$ is given in Figure 3. Since $ACF$ is a two-force member, the reaction at $F$ is directed along $CF$.

Conclusion:

From the free-body diagram in Figure 3, write the expression connecting the components of the reaction $F$ with the horizontal and vertical separations between the points $F$ and $C$.

$\frac{F_x}{16\text{in}\text{.}}=\frac{F_y}{7\text{in}\text{.}}$ (VI)

Substitute $32\text{lb}$ for $F_x$ in equation (VI) and solve for $F_y$.

$\begin{array}{c}\frac{32\text{lb}}{16\text{in}\text{.}}=\frac{F_y}{7\text{in}\text{.}}\\ F_y=\left(\frac{7\text{in}\text{.}}{16\text{in}\text{.}}\right)32\text{lb}\\ F_y=14\text{lb}\\ F_y=14.0\text{lb}\uparrow \end{array}$

Substitute $14\text{lb}$ for $F_y$ in equation (IV) and solver for $B_y$.

$\begin{array}{c}{B}_y+14\text{lb}-48\text{lb}=0\\ B_y=34\text{lb}\\ B_y=34.0\text{lb}\uparrow \end{array}$

Therefore, the components of reaction at $B$ and $F$ if the $48\text{lb}$ load is applied at $D$ are $\underset{\_}{B_x=32.0\text{lb}\leftarrow }$, $\underset{\_}{B_y=34.0\text{lb}\uparrow }$, $\underset{\_}{F_x=32.0\text{lb}\to }$, and $\underset{\_}{F_y=14.0\text{lb}\uparrow }$.

To determine

The components of reaction at $B$ and $F$ if the $48\text{lb}$ load is applied at $E$.

Answer to Problem 6.88P

The components of reaction at $B$ and $F$ if the $48\text{lb}$ load is applied at $E$ are $\underset{\_}{B_x=32.0\text{lb}\to }$, $\underset{\_}{B_y=10.0\text{lb}\uparrow }$, $\underset{\_}{F_x=32.0\text{lb}\leftarrow }$, and $\underset{\_}{F_y=38.0\text{lb}\uparrow }$.

Explanation of Solution

The $x$ components of the forces at $B$ and $F$ are obtained from the free-body diagram given in Figure 1.

$B_x=32.0\text{lb}\leftarrow$

$F_x=32.0\text{lb}\to$

The load is applied at $E$. The free-body diagram of the member $CDB$ is given in Figure 4.

Conclusion:

The free-body diagram of the member $CDB$ when the load is applied at $E$ is exactly similar to its free-body diagram when the load is applied at $A$. So, it is obvious that the $y$ components of the reactions $B$ and $F$ are identical for both situations.

$B_y=10.0\text{lb}\uparrow$

$F_y=38.0\text{lb}\uparrow$

Therefore, the components of reaction at $B$ and $F$ if the $48\text{lb}$ load is applied at $E$ are $\underset{\_}{B_x=32.0\text{lb}\to }$, $\underset{\_}{B_y=10.0\text{lb}\uparrow }$, $\underset{\_}{F_x=32.0\text{lb}\leftarrow }$, and $\underset{\_}{F_y=38.0\text{lb}\uparrow }$.