Chapter 4.1, Problem 4.20P

The ladder AB, of length L and weight W, can be raised by BC. Determine the tension T required to raise end B just off floor (a) in terms of W and θ, (b) if h = 8 ft, L = 10 ft, and W = lb.

Fig. P4.21

To determine

The tension in the cable AB shown in figure $\text{P4}\text{.19}$ assuming that $a=0.32\text{m}$.

Answer to Problem 4.20P

The tension in the cable AB shown in figure $\text{P4}\text{.19}$ is $1.500\text{kN}$ along the $x$ direction.

Explanation of Solution

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

The free body diagram is sketched below as figure 1.

Here, $T_x$ is the $x$ component of tension in the cable , $T_y$ is the $y$ component of tension in the cable , $C_x$ and $C_y$ are $x$ and $y$ component of reaction at $C$.

Write the expression for the moment at $C$

$M_C=F\times D_{\perp }$ (I)

Here, $M_C$ is the moment at $C$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $C$ and the point where force is experienced.

The moment at $C$ is due to the $x$ component of tension $T$ in the cable $AB$ and $240\text{N}$ acting at end of the bracket.

Thus, the complete expression of $M_A$ is

${\displaystyle \sum M_C}=T_x\left(0.32\text{m}\right)-240(0.4\text{m)}-240\text{N}\left(0.8\text{m}\right)$ (II)

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

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

${\displaystyle \sum M_C}=T_x\left(0.32\text{m}\right)-240(0.4\text{m)}-240\text{N}\left(0.8\text{m}\right)=0$ (III)

Write the expression for the ration of $T_x$ and $T_y$.

$\frac{T_x}{T_y}=\frac{T\cos \theta }{T\sin \theta }$

$\frac{T_x}{T_y}=\tan \theta$ (IV)

Write the expression for $\tan \theta$.

$\tan \theta =\frac{0.24}{0.32}$

Write the expression for the magnitude of tension in the cable $T$.

$T=\sqrt{T_x^2+T_y^2}$ (V)

Here, $T$ is the magnitude of tension at $B$.

Calculation:

Rearrange equation (III) to get $T_x$.

$\begin{array}{l}{T}_x\left(0.32\text{m}\right)-240(0.4\text{m)}-240\text{N}\left(0.8\text{m}\right)=0\\ T_x=900\text{N}\end{array}$

Substitute $\frac{0.24}{0.32}$ for $\tan \theta$ in equation (IV) to get $\frac{T_x}{T_y}$.

$\frac{T_x}{T_y}=\frac{0.24}{0.32}$

Substitute $900\text{N}$ for $T_x$ in above Equation to get $T_y$.

$\begin{array}{c}\frac{900\text{N}}{T_y}=\frac{0.24}{0.32}\\ T_y=\frac{4}{3}\left(900\text{N}\right)\\ =1200\text{N}\end{array}$

Substitute $1600\text{N}$ for $T_x$ and $1200\text{N}$ for $T_y$ in equation (V) to get $T$.

$\begin{array}{c}T=\sqrt{{\left(900\text{N}\right)}^2+{\left(1200\text{N}\right)}^2}\\ T=1500\text{N}\times \frac{1\text{kN}}{1000\text{N}}\\ =1.500\text{kN}\end{array}$ (X)

Therefore, the tension in the cable AB shown in figure $\text{P4}\text{.19}$ assuming $a=0.32\text{m}$ is $1.500\text{kN}$ along positive $x$ direction.

To determine

The reaction at $C$ shown in figure $\text{P4}\text{.19}$ assuming $a=0.32\text{m}$.

Answer to Problem 4.20P

The reaction at $C$ shown in figure $\text{P4}\text{.19}$ is $1.906\text{kN}$ and it acts at angle of $61.8°$ above the positive $x$ axis.

Explanation of Solution

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

${\displaystyle \sum F_x}=-T_x+C_x$ (VI)

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

The negative sign for $T_x$ implies that direction of $T_x$ is opposite to the direction of $C_x$ , which is in the $x$ direction.

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

${\displaystyle \sum F_x}=-T_x+C_x=0$ (VII)

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

${\displaystyle \sum F_y}=-T_y+C_y-240\text{N}-\text{240}\text{N}$ (VIII)

Except the force $C_y$ , the force $T_y$ and $240\text{N}$ act in the negative $y$ direction.

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

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

${\displaystyle \sum F_y}=-T_y+C_y-240\text{N}-\text{240}\text{N=0}$ (IX)

Write the expression for the magnitude of reaction at $C$.

$C=\sqrt{C_x^2+C_y^2}$ (X)

Here, $T$ is the magnitude of tension at $B$.

Calculation:

Substitute $900\text{N}$ for $T_x$ and rearrange the equation (VII) to get $C_x$.

$\begin{array}{l}-1600\text{N}+C_x=0\\ C_x=900\text{N}\end{array}$

Substitute $1200\text{N}$ for $T_y$ and rearrange the equation (IX) to get $C_y$.

$\begin{array}{l}-1200\text{N}+C_y-240\text{N}-\text{240}\text{N=0}\\ {\text{C}}_y=1680\text{N}\end{array}$

Substitute $900\text{N}$ for $C_x$ and $1680\text{N}$ for $C_y$ in equation (X) to get $C$.

$\begin{array}{c}C=\sqrt{900{\text{N}}^2+1680{\text{N}}^2}\\ =1906\text{N}\times \frac{1\text{kN}}{1000\text{N}}\\ =1.906\text{kN}\end{array}$

Let $\alpha$ is the angle between $C$ and $C_x$ , then write the expression for $\tan \alpha$.

$\begin{array}{l}\tan \alpha =\frac{1680\text{N}}{900\text{N}}\\ \alpha =61.8°\end{array}$

Therefore, the tension in the cable AB shown in figure $\text{P4}\text{.19}$ is $1.906\text{kN}$ at an angle $61.8°$ above positive $x$ axis.