Chapter 2.3, Problem 2.50P

Two cables are tied together at C and are loaded as shown. Determine the range of values of P for which both cables remain taut.

Fig. P2.49 and P2.50

To determine

The range of values of the force $P$ for which the cables $CA$ and $CB$ remain taut.

Answer to Problem 2.50P

The range of values of the force $P$ for which the cables $CA$ and $CB$ remain taut is $\underset{\_}{179.3\text{N}<P<669\text{N}}$.

Explanation of Solution

In the arrangement in Fig. P2.50, the forces acting at the point $c$ are one of magnitude $200\text{N}$ and the other of unknown magnitude $P$.

The free body diagram corresponding to the given arrangement is shown in Figure 1.

When the system is in equilibrium, the total forces acting on the system will be zero. Moreover, the equilibrium condition imply that the components of the forces along $x$ direction and $y$ direction will be separately equal to zero.

Write the expression for the equilibrium state along $x$ direction and $y$ direction.

$\Sigma F_x=0$ (I)

$\Sigma F_y=0$ (II)

Here, $F_x$ and $F_y$ are the components of the forces along $x$ and $y$ direction respectively.

Apply the condition in equation (I) to the given system.

$-T_{CA}\sin 30°+T_{CB}\sin 30°+P\cos 45°-200\text{N}=0$ (III)

Apply the condition in equation (II) to the given system.

$T_{CA}\cos 30°+T_{CB}\cos 30°-P\sin 45°=0$ (IV)

Write the condition for the cable $CA$ to remain taut.

$T_{CA}>0$ (V)

Write the condition for the cable $CA$ to remain taut.

$T_{CB}>0$ (VI)

The condition for both the cables to remain taut is that, there should be a tension greater than zero in each of the cables.

Conclusion:

Set $T_{CA}=0$ in equation (III) and (IV).

$\begin{array}{c}-\left(0\right)\sin 30°+T_{CB}\sin 30°+P\cos 45°-200\text{N}=0\\ T_{CB}\sin 30°+P\cos 45°-200\text{N}=0\end{array}$ (VII)

$\begin{array}{c}\left(0\right)\cos 30°+T_{CB}\cos 30°-P\sin 45°=0\\ T_{CB}\cos 30°-P\sin 45°=0\end{array}$ (VIII)

Add equation (VII) and (VIII).

$\begin{array}{c}{T}_{CB}\left(\sin 30°+\cos 30°\right)+P\left(\sin 45°-\cos 45°\right)-200\text{N}=0\\ T_{CB}\left(\sin 30°+\cos 30°\right)-200\text{N}=0\end{array}$ (IX)

Solve equation (IX) for $T_{CB}$.

$\begin{array}{c}{T}_{CB}=\frac{200\text{N}}{\left(\sin 30°+\cos 30°\right)}\\ =146.41\text{N}\end{array}$

Solve equation (VIII) for $P$ and substitute $146.41\text{N}$ for $T_{CB}$ to find the magnitude of $P$.

$\begin{array}{c}P=\frac{T_{CB}\cos 30°}{\sin 45°}\\ =\frac{\left(146.41\text{N}\right)\cos 30°}{\sin 45°}\\ =179.3\text{N}\end{array}$ (X)

Set $T_{CB}=0$ in equation (III) and (IV).

$\begin{array}{c}-T_{CA}\sin 30°+\left(0\right)\sin 30°+P\cos 45°-200\text{N}=0\\ -T_{CA}\sin 30°+P\cos 45°-200\text{N}=0\end{array}$ (XI)

$\begin{array}{c}{T}_{CA}\cos 30°+\left(0\right)\cos 30°-P\sin 45°=0\\ T_{CA}\cos 30°-P\sin 45°=0\end{array}$ (XII)

Add equation (XI) and (XII).

$\begin{array}{c}{T}_{CA}\left(\cos 30°-\sin 30°\right)+P\left(\cos 45°-\sin 45°\right)-200\text{N}=0\\ T_{CA}\left(\cos 30°-\sin 30°\right)-200\text{N}=0\end{array}$ (XIII)

Solve equation (XIII) for $T_{CA}$.

$\begin{array}{c}{T}_{CA}=\frac{200\text{N}}{\left(\cos 30°-\sin 30°\right)}\\ =546.41\text{N}\end{array}$

Solve equation (XII) for $P$ and substitute $546.41\text{N}$ for $T_{CA}$ to find the magnitude of $P$.

$\begin{array}{c}P=\frac{T_{CA}\cos 30°}{\sin 45°}\\ =\frac{\left(546.41\text{N}\right)\cos 30°}{\sin 45°}\\ =669.2\text{N}\end{array}$ (XIV)

The magnitude of $P$ obtained in equation (X) and (XIV) are the range of values of $P$ for which both the cables remain taut.

Therefore, the range of values of the force $P$ for which the cables $CA$ and $CB$ remain taut is $\underset{\_}{179.3\text{N}<P<669.2\text{N}}$.