Chapter 4.1, Problem 4.27P

For the frame and loading shown, determine the reactions at A E when (a) α = 30º, (b) α = 45 º.

Fig. P4.27

To determine

The reaction at $A$ and $E$ when $\alpha =30°$.

Answer to Problem 4.27P

The reaction at $A$ is $8.29\text{lb}$ and it acts angle $58.0°$ below positive $x$ axis and reaction at $E$ is $31.2\text{lb}$ and it acts at angle of $60.0°$ above positive $x$ axis.

Explanation of Solution

Forces acting upward and rightward are considered as positive and the torque acting counter clockwise is considered as positive.

Let $A$ is the reaction at the point $A$ shown in figure $\text{P4}\text{.25}$, $A_x$ and $A_y$ are the $x$ component and $y$ component of reaction at $A$. Let $E$ is the reaction at point $E$.

The free body diagram is sketched below as figure 1.

Here, $A$ is the magnitude of reaction at point $A$ , $E$ is the magnitude of reaction at point $E$ , $\alpha$ is the angle that reaction $E$ makes with $y$ axis, $A_x$ is the magnitude of the $x$ component of reaction at $A$ and $A_y$ is the magnitude of $y$ component of reaction at $A$.

Write the expression for the moment at $A$

${\displaystyle \sum M_A}={\displaystyle \sum F\times D_{\perp }}$ (I)

Here, ${\displaystyle \sum M_A}$ is the net moment at $A$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $A$ and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force about that point.

The moment at $A$ is due to the $x$ and $y$ component of reaction $E$ , $20\text{lb}$ force at $B$ and $20\text{lb}$ force at $C$.

The angle $\alpha$ is equal to $30°$.

The complete expression of moment $M_A$ at point $A$.

${\displaystyle \sum M_A}=E_x\left(8\text{in}\text{.}\right)+E_y\left(5\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(10\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(3\text{in}\text{.}\right)$ (II)

Here, ${\displaystyle \sum M_A}$ is the sum of all moment of force at $A$ , $E_x$ and $E_y$ are the $x$ and y component of $E$.

From figure 1, write the expression for the $x$ component of reaction $E$.

$E_x=E\sin \alpha$

From figure 1, write the expression for the $y$ component of reaction $E$.

$E_y=E\cos \alpha$

At equilibrium, the sum of the moment acting at $A$ will be zero.

Write the expression for the total moment acting at $A$.

${\displaystyle \sum M_A}=E\sin \alpha \left(8\text{in}\text{.}\right)+E\cos \alpha \left(5\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(10\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(3\text{in}\text{.}\right)=0$ (III)

The forces along the $x$ direction are the $x$ component of reaction $A$, $20\text{lb}$ force acting at $C$ an d the $x$ component of reaction $E$.

Therefore, write the expression for the net force along the $x$ direction.

${\displaystyle \sum F_x}=A_x-20\text{lb+}E\sin \alpha$ (IV)

Here, ${\displaystyle \sum F_x}$ is the sum of all force along the $x$ direction.

At equilibrium, the net force along the $x$ direction will be zero.

${\displaystyle \sum F_x}=A_x-20\text{lb+}E\sin \alpha =0$ (V)

The forces along the $x$ direction are the $y$ component of reaction $A$, $20\text{lb}$ force acting at $B$ and the $y$ component of reaction $E$.

Therefore, write the expression for the net force along the $y$ direction.

${\displaystyle \sum F_y}=A_y-20\text{lb}+E\cos \alpha$ (VI)

Here, ${\displaystyle \sum F_y}$ is the sum of all force along the $y$ direction.

At equilibrium, the net force along the $y$ direction will be zero.

${\displaystyle \sum F_y}=A_y-20\text{lb}+E\cos \alpha =0$ (VII)

Write the expression for the magnitude of net reaction at $A$.

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

Here, $A$ is the magnitude of net reaction at $A$.

Let $\theta$ be the angle between $A$ and $x$ axis.

Therefore, write the expression for the $\tan \theta$.

$\tan \theta =\frac{A_y}{A_x}$ (IX)

Calculation:

Substitute $30°$ for $\alpha$ and rearrange the equation (III) to get $E$.

$E\sin \left(30°\right)\left(8\text{in}\text{.}\right)+E\cos \left(30°\right)\left(5\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(10\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(3\text{in}\text{.}\right)=0$

$\begin{array}{c}E=\frac{\left(20\text{lb}\right)\left(10\text{in}\text{.}\right)+\left(20\text{lb}\right)\left(3\text{in}\text{.}\right)}{\sin \left(30°\right)\left(8\text{in}\text{.}\right)+\cos \left(30°\right)\left(5\text{in}\text{.}\right)}\\ =31.212\text{lb}\\ =3\text{1}\text{.2}\text{lb}\end{array}$

Substitute $30°$ for $\alpha$ and $31.212\text{lb}$ for $E$ in equation (V) to get $A_x$.

$A_x-20\text{lb+}\left(31.212\text{lb}\right)\sin \left(30°\right)=0$

$\begin{array}{c}{A}_x=20\text{lb}-\left(31.212\text{lb}\right)\sin \left(30°\right)\\ =\text{4}\text{.394}\text{lb}\end{array}$

Substitute $30°$ for $\alpha$ and $31.212\text{lb}$ for $E$ in equation (VII) to get $A_y$.

$A_y-20\text{lb}+\left(31.212\text{lb}\right)\cos \left(30°\right)=0$

$\begin{array}{c}{A}_y=20\text{lb}-\left(31.212\text{lb}\right)\cos \left(30°\right)\\ =-7.03\text{lb}\end{array}$

The negative sign indicates that direction of $A_y$ is along $-y$ axis.

Substitute $4.394\text{lb}$ for $A_x$ and $-7.03\text{lb}$ for $A_y$ in equation (VIII) to get $A$.

$\begin{array}{c}A=\sqrt{{\left(4.394\text{lb}\right)}^2+{\left(-7.03\text{lb}\right)}^2}\\ =8.29\text{lb}\end{array}$

Substitute $4.394\text{lb}$ for $A_x$ and $-7.03\text{lb}$ for $A_y$ in equation (IX) to get $\theta$.

$\begin{array}{c}\tan \theta =\frac{-7.03\text{lb}}{4.394\text{lb}}\\ \theta =-58.0°\end{array}$

The negative sign indicates that $A$ makes $58.0°$ below positive $x$ axis.

Therefore, the reaction at $A$ is $8.29\text{lb}$ and it acts angle $58.0°$ below positive $x$ axis and reaction at $E$ is $31.2\text{lb}$ and it acts at angle of $60.0°$ above positive $x$ axis.

To determine

The reaction at $A$ and $E$ when $\alpha =45°$.

Answer to Problem 4.27P

The reaction at $A$ is $0\text{lb}$ and reaction at $E$ is $28.3\text{lb}$ and it acts at angle of $45.0°$ above positive $x$ axis.

Explanation of Solution

Take all vectors along the $x$ axis and $y$ axis as positive.

Let $A$ is the reaction at the point $A$ shown in figure $\text{P4}\text{.27}$, $A_x$ and $A_y$ are the $x$ component and $y$ component of reaction at $A$. Let $E$ is the reaction at point $E$.

The free body diagram is sketched in figure 1.

Write the expression for the moment at $A$

${\displaystyle \sum M_A}={\displaystyle \sum F\times D_{\perp }}$ (I)

Here, ${\displaystyle \sum M_A}$ is the net moment at $A$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $A$ and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at $A$ is due to the $x$ and $y$ component of reaction $E$ , $20\text{lb}$ force at $B$ and $20\text{lb}$ force at $C$.

The angle $\alpha$ is equal to $45°$.

Thus, the complete expression of net anticlockwise moment $M_A$ at point $A$.

${\displaystyle \sum M_A}=E_x\left(8\text{in}\text{.}\right)+E_y\left(5\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(10\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(3\text{in}\text{.}\right)$ (II)

Here, ${\displaystyle \sum M_A}$ is the sum of all moment of force at $A$ , $E_x$ and $E_y$ are the $x$ and y component of $E$.

From figure 1, write the expression for the $x$ component of reaction $E$.

$E_x=E\sin \alpha$

From figure 1, write the expression for the $y$ component of reaction $E$.

$E_y=E\cos \alpha$

At equilibrium, the sum of the moment acting at $A$ will be zero.

Write the expression for the total moment acting at $A$.

${\displaystyle \sum M_A}=E\sin \alpha \left(8\text{in}\text{.}\right)+E\cos \alpha \left(5\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(10\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(3\text{in}\text{.}\right)=0$ (III)

The forces along the $x$ direction are the $x$ component of reaction $A$, $20\text{lb}$ force acting at $C$ an d the $x$ component of reaction $E$.

Therefore, write the expression for the net force along the $x$ direction.

${\displaystyle \sum F_x}=A_x-20\text{lb+}E\sin \alpha$ (IV)

Here, ${\displaystyle \sum F_x}$ is the sum of all force along the $x$ direction.

At equilibrium, the net force along the $x$ direction will be zero.

${\displaystyle \sum F_x}=A_x-20\text{lb+}E\sin \alpha =0$ (V)

The forces along the $x$ direction are the $y$ component of reaction $A$, $20\text{lb}$ force acting at $B$ and the $y$ component of reaction $E$.

Therefore, write the expression for the net force along the $y$ direction.

${\displaystyle \sum F_y}=A_y-20\text{lb}+E\cos \alpha$ (VI)

Here, ${\displaystyle \sum F_y}$ is the sum of all force along the $y$ direction.

At equilibrium, the net force along the $y$ direction will be zero.

${\displaystyle \sum F_y}=A_y-20\text{lb}+E\cos \alpha =0$ (VII)

Write the expression for the magnitude of net reaction at $A$.

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

Here, $A$ is the magnitude of net reaction at $A$.

Let $\theta$ be the angle between $A$ and $x$ axis.

Therefore, write the expression for the $\tan \theta$.

$\tan \theta =\frac{A_y}{A_x}$ (IX)

Calculation:

Substitute $45°$ for $\alpha$ and rearrange the equation (III) to get $E$.

$E\sin \left(45°\right)\left(8\text{in}\text{.}\right)+E\cos \left(45°\right)\left(5\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(10\text{in}\text{.}\right)-\left(20\text{lb}\right)\left(3\text{in}\text{.}\right)=0$

$\begin{array}{c}E=\frac{\left(20\text{lb}\right)\left(10\text{in}\text{.}\right)+\left(20\text{lb}\right)\left(3\text{in}\text{.}\right)}{\sin \left(45°\right)\left(8\text{in}\text{.}\right)+\cos \left(45°\right)\left(5\text{in}\text{.}\right)}\\ =28.28\text{lb}\\ \text{=28}\text{.3}\text{lb}\end{array}$

Substitute $45°$ for $\alpha$ and $28.28\text{lb}$ for $E$ in equation (V) to get $A_x$.

$A_x-20\text{lb+}\left(28.28\text{lb}\right)\sin \left(45°\right)=0$

$\begin{array}{c}{A}_x=20\text{lb}-\left(28.28\text{lb}\right)\sin \left(45°\right)\\ =\text{0}\text{lb}\end{array}$

Substitute $45°$ for $\alpha$ and $28.28\text{lb}$ for $E$ in equation (VII) to get $A_y$.

$A_y-20\text{lb}+\left(28.28\text{lb}\right)\cos \left(45°\right)=0$

$\begin{array}{c}{A}_y=20\text{lb}-\left(28.28\text{lb}\right)\cos \left(45°\right)\\ =0\text{lb}\end{array}$

The negative sign indicates that direction of $A_y$ is along $-y$ axis.

Substitute $0\text{lb}$ for $A_x$ and $0\text{lb}$ for $A_y$ in equation (VIII) to get $A$.

$\begin{array}{c}A=\sqrt{{\left(0\text{lb}\right)}^2+{\left(0\text{lb}\right)}^2}\\ =0\text{lb}\end{array}$

Therefore, the reaction at $A$ is $0\text{lb}$ and reaction at $E$ is $28.3\text{lb}$ and it acts at angle of $45.0°$ above positive $x$ axis.