Chapter 4.1, Problem 4.23P

4.23 and 4.24 For each of the plates and loadings shown, determine the reaction at A and B.

Fig. P4.23

Fig. P4.24

To determine

The reaction at $A$ and $B$ of the plate shown in $\text{P4}\text{.23(a)}$.

Answer to Problem 4.23P

The reaction at $A$ is $44.7\text{lb}$ and is directed along $26.6°$ above negative $x$ axis and reaction at $B$ is $30.0\text{lb}$ in the upward direction.

Explanation of Solution

Take vectors along positive $x$ and $y$ axis are positive.

Let $A$ is the reaction at the point $A$ , $B$ is the reaction at the point $B$ and $A_x$ and $A_y$ are the component of the reaction $A$.

The free body diagram is sketched below as figure 1.

Here, $A_x$ and $A_y$ are the magnitude of $x$ and $y$ component of reaction $A$ at the point $A$ and $B$ is the magnitude of reaction $B$.

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 reaction $B$ and forces $50\text{lb}$ and $40\text{lb}$.

Thus, the complete expression of $M_A$ is

${\displaystyle \sum M_A}=B\left(20\text{in}\text{.}\right)-\left(50\text{lb}\right)(4\text{in}\text{.)}-\left(40\text{lb}\right)\left(10\text{in}\text{.}\right)$ (II)

Here, ${\displaystyle \sum M_A}$ is the sum of all moment of force at $A$.

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

${\displaystyle \sum M_A}=B\left(20\text{in}\text{.}\right)-\left(50\text{lb}\right)(4\text{in}\text{.)}-\left(40\text{lb}\right)\left(10\text{in}\text{.}\right)=0$ (III)

Write the expression for the net force along the $x$ direction.

${\displaystyle \sum F_x}=A_x+40\text{lb}$ (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.

${\displaystyle \sum F_x}=A_x+40\text{lb}=0$ (V)

Write the expression for the net force along the $y$ direction.

${\displaystyle \sum F_y}=A_y+B-50\text{lb}\text{=0}$ (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.

${\displaystyle \sum F_y}=A_y+B-50\text{lb}\text{=0}$ (VII)

Let $\alpha$ be the angle that $A$ makes with $x$ axis.

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 $A$.

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

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

Calculation:

Rearrange equation (III) to get $B$.

$\begin{array}{l}B\left(20\text{in}\text{.}\right)-\left(50\text{lb}\right)(4\text{in}\text{.)}-\left(40\text{lb}\right)\left(10\text{in}\text{.}\right)=0\\ B=30\text{lb}\end{array}$

From figure 1, the reaction $B$ acts in the perpendicular direction.

Rearrange equation (V) to get $A_x$ .

$\begin{array}{l}{A}_x+40\text{lb}=0\\ A_x=-40\text{lb}\end{array}$

The negative sign indicates that the $A_x$ acts in opposite direction to that assumed initially. Therefore $A_x$ acts in the negative $x$ direction.

Rearrange equation (VII) to get $A_y$.

$A_y\text{=}50\text{lb}-B$

Substitute $30.0\text{lb}$ for $B$ in above equation to get $A_y$.

$\begin{array}{c}{A}_y\text{=}50\text{lb}-30.0\text{lb}\\ \text{=20}\text{lb}\end{array}$

Substitute $-40\text{lb}$ for $A_x$ and $20\text{lb}$ for $A_y$ in the equation (VIII) to get $A$.

$\begin{array}{c}A=\sqrt{{\left(-40\text{lb}\right)}^2+{\left(20\text{lb}\right)}^2}\\ =44.7\text{lb}\end{array}$

Substitute $-40\text{lb}$ for $A_x$ and $20\text{lb}$ for $A_y$ in the equation (IX) to get $\alpha$.

$\begin{array}{c}\tan \alpha =\frac{20\text{lb}}{-40\text{lb}}\\ \alpha =-26.6°\end{array}$

Therefore, the reaction at $A$ is $44.7\text{lb}$ and is directed along $26.6°$ above negative $x$ axis and reaction at $B$ is $30.0\text{lb}$ in the upward direction.

To determine

The reaction at $A$ and $B$ of the plate shown in $\text{P4}\text{.23(b)}$.

Answer to Problem 4.23P

The reaction at $A$ is $30.2\text{lb}$ and is directed $41.4°$ above negative $x$ axis and reaction at $B$ is $34.6\text{lb}$ and is directed $60.0°$ above negative $x$ axis

Explanation of Solution

Take vectors along positive $x$ and $y$ axis are positive.

Let $A$ is the reaction at the point $A$ , $B$ is the reaction at the point $B$ and $A_x$ and $A_y$ are the component of the reaction $A$.

The free body diagram is sketched below as figure 2.

Here, $A_x$ and $A_y$ are the magnitude of $x$ and $y$ component of reaction $A$ at the point $A$ and $B$ is the magnitude of reaction $B$.

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 $y$ component of the reaction $B$ and forces $50\text{lb}$ and $40\text{lb}$.

Thus, the complete expression of $M_A$ is

${\displaystyle \sum M_A}=\left(B\cos 30°\right)\left(20\text{in}\text{.}\right)-\left(50\text{lb}\right)(4\text{in}\text{.)}-\left(40\text{lb}\right)\left(10\text{in}\text{.}\right)$ (II)

Here, ${\displaystyle \sum M_A}$ is the sum of all moment of force at $A$.

From figure 2 , u component of reaction $B$ is $B\cos 30°$.

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

${\displaystyle \sum M_A}=\left(B\cos 30°\right)\left(20\text{in}\text{.}\right)-\left(50\text{lb}\right)(4\text{in}\text{.)}-\left(40\text{lb}\right)\left(10\text{in}\text{.}\right)=0$ (III)

Write the expression for the net force along the $x$ direction.

${\displaystyle \sum F_x}=A_x-B\sin 30°+40\text{lb}$ (IV)

Here, ${\displaystyle \sum F_x}$ is the sum of all force along the $x$ direction, $B\sin 30°$ is the component of the reaction B along $-x$ direction.

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

${\displaystyle \sum F_x}=A_x-B\sin 30°+40\text{lb}=0$ (V)

Write the expression for the net force along the $y$ direction.

${\displaystyle \sum F_y}=A_y+B\cos 30°-50\text{lb}$ (VI)

Here, ${\displaystyle \sum F_y}$ is the sum of all force along the $y$ direction, $B\cos 30°$ is the component of the reaction B along $y$ direction.

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

${\displaystyle \sum F_y}=A_y+B\cos 30°-50\text{lb}\text{=0}$ (VII)

Let $\alpha$ be the angle that $A$ makes with $x$ axis.

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 $A$.

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

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

Calculation:

Rearrange equation (III) to get $B$.

$\begin{array}{l}\left(B\cos 30°\right)\left(20\text{in}\text{.}\right)-\left(50\text{lb}\right)(4\text{in}\text{.)}-\left(40\text{lb}\right)\left(10\text{in}\text{.}\right)=0\\ B=34.64\text{lb}\end{array}$

From figure 2, the reaction B is $60°$ above negative $x$ axis.

Rearrange equation (V) to get $A_x$ .

$A_x=B\sin 30°-40\text{lb}$

Substitute $34.64\text{lb}$ for $B$ in the above equation to get $A_x$.

$\begin{array}{c}{A}_x=\left(34.64\text{lb}\right)\sin 30°-40\text{lb}\\ \text{=}-\text{22}\text{.68}\text{lb}\end{array}$

$A_x$ acts in the $-x$ direction.

Rearrange equation (VII) to get $A_y$.

$A_y\text{=}50\text{lb}-B\cos 30°$

Substitute $34.64\text{lb}$ for $B$ in the above equation to get $A_y$.

$\begin{array}{c}{A}_y=50\text{lb}-\left(34.64\right)\cos 30°\\ =20\text{lb}\end{array}$

Substitute $-\text{22}\text{.68}\text{lb}$ for $A_x$ and $20\text{lb}$ for $A_y$ in the equation (VIII) to get $A$.

$\begin{array}{c}A=\sqrt{{\left(-\text{22}\text{.68}\text{lb}\right)}^2+{\left(20\text{lb}\right)}^2}\\ =30.23\text{lb}\end{array}$

Substitute $-\text{22}\text{.68}\text{lb}$ for $A_x$ and $20\text{lb}$ for $A_y$ in the equation (IX) to get $\alpha$.

$\begin{array}{c}\tan \alpha =\frac{20\text{lb}}{-\text{22}\text{.68}\text{lb}}\\ \alpha =-41.4°\end{array}$

Therefore, the reaction at $A$ is $30.2\text{lb}$ and is directed $41.4°$ above negative $x$ axis and reaction at $B$ is $34.6\text{lb}$ and is directed $60.0°$ above negative $x$ axis