Chapter 5, Problem 28SP

The uniform bar in Fig. 5-18 weighs 40 N and is subjected to the forces shown. Find the magnitude, location, and direction of the force needed to keep the bar in equilibrium.

Fig. 5-18

To determine

The magnitude, location, and direction of the force to keep the bar, in figure $5.18$, in equilibriumif theweight of the uniform bar is $40\text{ N}$.

Answer to Problem 28SP

Solution:

$\text{0}\text{.11 kN}$, $\text{0}\text{.68 L}$, and $49°$

Explanation of Solution

Given data:

The weight of the uniform horizontal bar is $\text{40 N}$.

Formula used:

Write the expression for the first condition of force equilibrium:

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

Here, ${\displaystyle \sum F_x}$ is the sum of the forces in the x- direction (horizontal direction) and ${\displaystyle \sum F_y}$ is the sum of the forces in the y-direction (vertical direction).

From the parallelogram law of vector addition, the expression for the resultant force is

$\overrightarrow{R}=\left(F_x\widehat{i}+F_y\widehat{j}\right)$

Magnitude of $\overrightarrow{R}$ is

$\overrightarrow{R}=\sqrt{{\left(F_x\right)}^2+{\left(F_y\right)}^2+2F_x{F}_y\cos \varphi }$

Here, $\varphi$ is the angle between both the vectors.

Direction of $\overrightarrow{R}$ is

$\theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)$

Here, $F_x$ is the force in the x- direction or in the horizontal direction and $F_y$ is the force in the y-direction or in the vertical direction.

Write the expression for torque:

$\tau =rF\sin \theta$

Here, $r$ is the distance of the applied force, $F$, from the axis of rotation and $\theta$ is the angle.

Write the expression for the torque equilibrium condition:

${\displaystyle \sum {\tau }_O}=0$

Here, ${\displaystyle \sum {\tau }_O}$ is the sum of the torque about point $O$.

Explanation:

Draw the free body diagram of the uniform horizontal bar:

In the above diagram, at point $P$, $\overrightarrow{R}$ is the resultant of the force, $\overrightarrow{R}$ has two forcecomponents, $F_x$ is the horizontal force at point $P$ along the horizontal direction, $F_y$ is the vertical force at point P along the vertical direction, $x$ is the distance of resultent force from the point $O$, $80\cos 30°$ and $80\sin 30°$ are the orizontal and the vertical force components of $80\text{ N}$ respectively, and $40\text{ N}$ is the weight of the horizontal bar.

Recall the expression for the first condition of forces equilibrium along the horizontal direction:

${{\displaystyle \sum F}}_x=0$

Consider the direction of the rightward forces as positive and the direction of the leftward forces as negative. Therefore,

$\begin{array}{c}-F_x-80\text{ N}\left(\cos 30°\right)=0\\ F_x=-69.28\text{ N}\end{array}$

Recall the expression for the first condition of forces equilibrium along the vertical direction:

${{\displaystyle \sum F}}_y=0$

Consider the direction of the upward forces as positive and the direction of the downward forces as negative. Therefore,

$\begin{array}{c}{F}_y+50\text{ N+80 N}\left(\sin 30°\right)-60\text{ N}-40\text{ N}-70\text{ N}=0\\ F_y\text{=}-80\text{ N }\end{array}$

Recall the expression formagnitude of the resultant force:

$R=\sqrt{{\left(F_x\right)}^2+{\left(F_y\right)}^2+2F_x{F}_y\cos \varphi }$

Substitute $-69.28\text{ N}$ for $F_x$, $-80\text{ N }$ for $F_y$, and $90°$ for $\varphi$

$\begin{array}{c}\overrightarrow{R}=\sqrt{{\left(-69.28\text{ N}\right)}^2+{\left(-80\text{ N}\right)}^2+\left(-69.28\text{ N}\right)\left(-80\text{ N}\right)\cos 90°}\\ =\sqrt{{\left(-69.28\text{ N}\right)}^2+{\left(-80\text{ N}\right)}^2}\\ =106\text{ N}\left(\frac{1\text{ kN}}{1000\text{ N}}\right)\\ \simeq 0.11\text{ kN}\end{array}$

Recall the expression for the direction of the resultant force:

$\theta ={\tan }^{-1}\left(\frac{F_y}{F_x}\right)$

Substitute $-69.28\text{ N}$ for $F_x$ and $-80\text{ N }$ for $F_y$

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-80\text{ N }}{-69.28\text{ N}}\right)\\ ={\tan }^{-1}\left(1.154\right)\\ =49°\end{array}$

Find the location of the force from the right-hand side at point $O$:

Recall the expression for $\tau$

$\tau =rF\sin \theta$

Apply the torque equation at point $O$ by considering the counter clockwise direction of the torque as positive and the clockwise direction as negative:

$\begin{array}{c}{\displaystyle \sum {\tau }_O}=0\\ \left[\begin{array}{l}\left(60\text{ N}\right)\left(L\right)\left(\sin 90°\right)-\left(50\text{ N}\right)\left(0.80L\right)\left(\sin 90°\right)\\ +\left(40\text{ N}\right)\left(0.50L\right)\left(\sin 90°\right)-\left(70\text{ N}\right)\left(0.20L\right)\left(\sin 90°\right)\end{array}\right]+F_y\left(x\right)=0\end{array}$

Substitute $-80\text{ N }$ for $F_y$

$\left[\begin{array}{l}\left(60\text{ N}\right)\left(L\right)\left(\sin 90°\right)-\left(50\text{ N}\right)\left(0.80L\right)\left(\sin 90°\right)\\ +\left(40\text{ N}\right)\left(0.50L\right)\left(\sin 90°\right)-\left(70\text{ N}\right)\left(0.20L\right)\left(\sin 90°\right)\end{array}\right]+\left(-80\text{ N}\right)\left(x\right)=0$

Solve the equation for $x$

$\begin{array}{c}60L-40L+20L+14L=80x\\ 54L=80x\\ x=0.675L\\ \simeq 0.68L\end{array}$

Conclusion:

The magnitude of the force is $\text{0}\text{.11 kN}$, the location of the force from the right-hand side is $\text{0}\text{.68 L}$, and the direction of the force is $49°$.