Chapter 4.1, Problem 4.27P

(a)

To determine

Find the reactions at A and B when $\alpha =0$.

Answer to Problem 4.27P

The reaction at A is $\underset{\_}{37.5\text{lb}}$.

The reaction at B is $\underset{\_}{37.5\text{lb}}$.

Explanation of Solution

Given information:

The value of angle is $\alpha =0$.

Assumption:

Apply the sign convention for calculating the equations of equilibrium as below:

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive.

Calculation:

Consider the condition $\alpha =0°$.

Draw the free body diagram when $\alpha =0°$ at B. At an angle $\alpha =0°$, the vertical force only acting at B.

Sketch the Free Body Diagram as shown in Figure 1.

Refer to Figure 1.

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium of moment about A is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ B_y\times 20-75\times 10=0\\ 20B_y=750\\ B_y=37.5\text{lb}\end{array}$

Hence, the reaction at B is $\underset{\_}{37.5\text{lb}}$.

Apply the Equilibrium of forces along x direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x=0\end{array}$

Apply the Equilibrium of forces along y direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y-75+37.5=0\\ A_y=37.5\text{lb}\end{array}$

Hence, the reaction at A is $\underset{\_}{37.5\text{lb}}$.

(b)

To determine

The reactions at A and B when $\alpha =90°$.

Answer to Problem 4.27P

The reaction at A is $\underset{\_}{97.6\text{lb}\nearrow \text{50}\text{.2}°}$.

The reaction at B is $\underset{\_}{62.5\text{lb}}$.

Explanation of Solution

Given information:

The angle is $\alpha =90°$.

Calculation:

Consider the condition $\alpha =90°$.

Draw the free body diagram when $\alpha =90°$ at B. At an angle $\alpha =90°$, the horizontal force only acting at B.

Sketch the Free Body Diagram as shown in Figure 2.

Refer to Figure 2.

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium of moment about A is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ B_x\times 12-75\times 10=0\\ 12B_x=750\\ B_x=62.5\text{lb}\end{array}$

Hence, the reaction at B is $\underset{\_}{62.5\text{lb}}$.

Apply the Equilibrium of forces along x direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x-B_x=0\end{array}$

Substitute $62.5\text{lb}$ for $B_x$.

$A_x=62.5\text{lb}$

Apply the Equilibrium of forces along y direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y-75=0\\ A_y=75\text{lb}\end{array}$

Calculate the resultant reaction at A as shown below.

$A=\sqrt{A_x^2+A_y^2}$

Substitute $62.5\text{lb}$ for $A_x$ and $75\text{lb}$ for $A_y$.

$\begin{array}{c}A=\sqrt{{62.5}^2+{75}^2}\\ =\sqrt{9,531.25}\\ =97.6\text{lb}\end{array}$

Calculate the angle $\left(\theta \right)$ as shown below.

$\tan \theta =\frac{A_y}{A_x}$

Substitute $62.5\text{lb}$ for $A_x$ and $75\text{lb}$ for $A_y$.

$\begin{array}{l}\tan \theta =\frac{75}{62.5}\\ \theta ={\tan }^{-1}\left(1.2\right)\\ \theta =50.2°\end{array}$

Hence, the reaction at A is $\underset{\_}{97.6\text{lb}\nearrow \text{50}\text{.2}°}$.

(c)

To determine

The reactions at A and B when $\alpha =30°$.

Answer to Problem 4.27P

The reaction at A is $\underset{\_}{49.8\text{lb}\nearrow \text{71}\text{.2}°}$.

The reaction at B is $\underset{\_}{32.2\text{lb}\nwarrow \text{60}°}$.

Explanation of Solution

Given information:

The angle is $\alpha =30°$.

Calculation:

Consider the condition $\alpha =30°$.

Draw the free body diagram when $\alpha =30°$ at B.

Sketch the Free Body Diagram as shown in Figure 3.

Refer to Figure 3.

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium of moment about A is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ \left(B\cos 30°\times 20\right)+\left(B\sin 30°\times 12\right)-\left(75\times 10\right)=0\\ 23.32B=750\\ B=32.16\text{lb}\end{array}$

Hence, the reaction at B is $\underset{\_}{32.16\text{lb}\nwarrow \text{60}°}$.

Apply the Equilibrium of forces along x direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x-B\sin 30°=0\end{array}$

Substitute $32.16\text{lb}$ for B.

$\begin{array}{l}{A}_x-32.16\sin 30°=0\\ A_x=16.08\text{lb}\end{array}$

Apply the Equilibrium of forces along y direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y-75+32.16\cos 30°=0\\ A_y=47.149\text{lb}\end{array}$

Calculate the resultant reaction at A as shown below.

$A=\sqrt{A_x^2+A_y^2}$

Substitute $16.08\text{lb}$ for $A_x$ and $47.149\text{lb}$ for $A_y$.

$\begin{array}{c}A=\sqrt{{16.08}^2+{47.149}^2}\\ =\sqrt{2,481.5946}\\ =49.8\text{lb}\end{array}$

Calculate the angle $\left(\alpha \right)$ as shown below.

$\tan \alpha =\frac{A_y}{A_x}$

Substitute $16.08\text{lb}$ for $A_x$ and $47.149\text{lb}$ for $A_y$.

$\begin{array}{l}\tan \alpha =\frac{47.149}{16.08}\\ \alpha ={\tan }^{-1}\left(2.93215\right)\\ \alpha =71.2°\end{array}$

Therefore, the reaction at A is $\underset{\_}{49.8\text{lb}\nearrow \text{71}\text{.2}°}$.