Chapter 12, Problem 12.66CP

In the What If? section of Example 12.2, let d represent the distance in meters between the person and the hinge at the left end of the beam. (a) Show that the cable tension is given by T = 93.9d + 125, with T in newtons. (b) Show that the direction angle θ of the hinge force is described by

$\tan \theta =\left(\frac{32}{3d+4}-1\right)\tan 53.0°$

(c) Show that the magnitude of the hinge force is given by

$R=\sqrt{8.82\times {10}^3{d}^2-9.65\times {10}^{4}d+4.96\times {10}^5}$

(d) Describe how the changes in T, θ, and R as d increases differ from one another.

To determine

The cable tension is $T=93.9d+125$.

Answer to Problem 12.66CP

The cable tension is given by, $T=93.9d+125$.

Explanation of Solution

The weight of the person is $600\text{N}$, the weight of the beam is $200\text{N}$, the length of the beam is $8\text{m}$, cables makes an angle of $53°$, distance between the person and hinged support is $d$, the tension in the rope is $T$.

The diagram for the given condition is shown below.

Figure (1)

Apply the rotational equilibrium equation and take the torque about hinge support.

    $\left(T\sin \varphi \right)\times l-w_{\text{p}}\times d-w_{\text{b}}\times \frac{l}{2}=0$        (1)

Here, $w_{\text{p}}$ is the weight of the person, $w_{\text{b}}$ is the weight of the beam, $l$ is the length of the beam an d $\varphi$ is the cable angle.

Rearrange the equation (1) for $T$.

    $T=\frac{w_{\text{p}}\times d+w_{\text{b}}\times \frac{l}{2}}{l\times \sin \varphi }$        (2)

Substitute, $600\text{N}$ for $w_{\text{p}}$, $200\text{N}$ for $w_{\text{b}}$, $8\text{m}$ for $l$ and $53°$ for $\varphi$ in equation (2) to find the expression for the tension.

    $\begin{array}{c}T=\frac{600\text{N}\times d+200\text{N}\times \frac{8}{2}}{8\times \sin 53}\\ T=93.9d+125\end{array}$

Conculasion:

Therefore, the cable tension will be $T=93.9d+125$.

To determine

The direction angle is $\tan \theta =\left(\frac{32}{3d+4}-1\right)\tan 53°$

Answer to Problem 12.66CP

The direction angle will be $\tan \theta =\left(\frac{32}{3d+4}-1\right)\tan 53°$.

Explanation of Solution

Apply the equilibrium condition for the horizontal forces.

    $\begin{array}{c}\sum F_{\text{x}}=0\\ R\cos \theta -T\cos \varphi =0\\ R\cos \theta =T\cos \varphi \end{array}$        (3)

Apply the equilibrium equation for the vertical forces.

    $\begin{array}{c}\sum F_{\text{y}}=0\\ R\sin \theta +T\sin \varphi -w_{\text{p}}-w_{\text{b}}=0\\ R\sin \theta =w_{\text{p}}+w_{\text{b}}-T\sin \varphi \end{array}$        (4)

Here, $R$ is the hinge reaction and $\theta$ is the angle makes by the hinge reaction.

Divide equation (4) by equation (3).

    $\begin{array}{c}\frac{R\sin \theta }{R\cos \theta }=\frac{w_{\text{p}}+w_{\text{b}}-T\sin \varphi }{T\cos \varphi }\\ \tan \theta =\left(\frac{w_{\text{p}}+w_{\text{b}}-T\sin \varphi }{T\cos \varphi }\right)\times \frac{\tan \varphi }{\tan \varphi }\\ \tan \theta =\left(\frac{w_{\text{p}}+w_{\text{b}}-T\sin \varphi }{T\cos \varphi }\right)\tan \varphi \\ \tan \theta =\left(\frac{w_{\text{p}}+w_{\text{b}}}{T\sin \varphi }-1\right)\tan \varphi \end{array}$        (5)

Substitute, $600\text{N}$ for $w_{\text{p}}$, $200\text{N}$ for $w_{\text{b}}$, $\left(93.9d+125\right)\text{N}$ for $T$  and $53°$ for $\varphi$ in equation (5) to find the reaction angle.

    $\begin{array}{c}\tan \theta =\left(\frac{600\text{N}+200\text{N}}{\left(93.9d+125\right)\text{N}\times \sin 53°}-1\right)\tan 53°\\ \tan \theta =\left(\frac{32}{3d+4}-1\right)\tan 53°\end{array}$

Conculasion:

Therefore, The direction angle will be $\tan \theta =\left(\frac{32}{3d+4}-1\right)\tan 53°$

To determine

The magnitude of the hinge force is $R=\sqrt{8.82\times {10}^5{d}^2-9.65\times {10}^{4}d+4.96\times {10}^5}$ .

Answer to Problem 12.66CP

The magnitude of the hinge force will be $R=\sqrt{8.82\times {10}^5{d}^2-9.65\times {10}^{4}d+4.96\times {10}^5}$.

Explanation of Solution

Square on both side of equation(2) and equation(3) and then add them.

    $\begin{array}{c}{R}^2\left({\cos }^2\theta +{\sin }^2\theta \right)=T^2{\cos }^2\varphi +{\left(w_{\text{p}}+w_{\text{b}}-T\sin \varphi \right)}^2\\ R=\sqrt{T^2+{\left(w_{\text{p}}+w_{\text{b}}\right)}^2-2T\sin \varphi \left(w_{\text{p}}+w_{\text{b}}\right)}\end{array}$        (6)

Substitute, $600\text{N}$ for $w_{\text{p}}$, $200\text{N}$ for $w_{\text{b}}$, $\left(93.9d+125\right)\text{N}$ for $T$  and $53°$ for $\varphi$ in equation (6) to find the reaction.

    $\begin{array}{c}R=\sqrt{{\left(93.9d+125\right)}^2\text{N}+{\left(600\text{N}+200\text{N}\right)}^2-2\left(93.9d+125\right)\text{N}\times \sin 53°\left(600\text{N}+200\text{N}\right)}\\ =\sqrt{8.82\times {10}^5{d}^2-9.65\times {10}^{4}d+4.96\times {10}^5}\end{array}$

Conclusion:

Therefore, the magnitude of the hinge force is $R=\sqrt{8.82\times {10}^5{d}^2-9.65\times {10}^{4}d+4.96\times {10}^5}$.

To determine

The changes in $T$, $\theta$ and $R$ on increasing the distance $d$.

Answer to Problem 12.66CP

$T$ and $R$ increases on increases the distance $d$ and $\theta$ decreases when the distance $d$ increases.

Explanation of Solution

$T$ and $R$ is the directely proportional to the distance d. since when $d$ is increases then $T$ and $R$ is increases. But angle $\theta$ is inversely proportional to the distance $d$ so when the $d$ is increases then the magnitude of the angle is decreases.

Conclusion:

Therefore, $T$ and $R$ increases on increases the distance $d$ and $\theta$ decreases.