Chapter 9, Problem 77QAP

To determine

(a)

The tension in the wire.

Answer to Problem 77QAP

The tension in the wire is $9.8\times {10}^{3}N$.

Explanation of Solution

Given:

Mass of the beam $m=1000kg$.

Let 'T' be the tension in the wire

Length of beam $d=4.0m$

Formula used:

Newton's 2^nd law: $F=ma$

Torque (or moment of force) $\tau =F\times d$

Here, all alphabets are in their usual meanings.

Calculation: Pictorial Diagram

  

Consider the pictorial diagram wherein all components are shown

Apply Newton's 2^nd law

For x − component:

  $\begin{array}{l}{\displaystyle \sum F_x=0}\\ F_x-T_x=0\\ F_x-T\cos 30°=0----(i)\end{array}$

For y − component:

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_y+T_y-mg=0\\ F_y+T_y-1000\times 9.8=0\\ F_y-T\sin 30°-9800=0----(ii)\end{array}$

Net Torque at the hinge ${\tau }_{F_y}=0and{\tau }_{F_x}=0$

Therefore, $mg\times \left(\frac{d}{2}\right)=\left(T\sin 30°\times d\right)$

  $\begin{array}{l}T=\frac{\left(mgd/2\right)}{\left(\sin 30°\times d\right)}\\ T=\frac{\left(1000\times 9.8\times 2.0\right)}{\left(\sin 30°\times 4.0\right)}N\\ T=\frac{\left(1000\times 9.8\times 2.0\right)}{\left(0.5\times 4.0\right)}N\\ T=9.8\times {10}^{3}N\end{array}$

Hence, the tension in the wire is $9.8\times {10}^{3}N$.

Conclusion:

Thus, the tension in the wire is $9.8\times {10}^{3}N$.

To determine

(b)

The minimum cross-sectional area of the wire.

Answer to Problem 77QAP

The minimum cross-sectional area of the wire is $2.45\times {10}^{-5}{m}^2$.

Explanation of Solution

Given:

Ultimate breaking strength $400\times {10}^{6}N/m^2$.

Let the minimum cross-sectional area of the wire be ' A'

Maximum applied force on the wire $F_{\max }=9.8\times {10}^{3}N$

Formula used:

Ultimate strength $=\frac{F_{\max }}{A_{\min }}$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in above formula,

  $Ultimatestrength=\frac{F_{\max }}{A_{\min }}$

  $\begin{array}{l}or,A_{\min }=\frac{F_{\max }}{\left(Ultimatestrength\right)}\\ or,A_{\min }=\frac{9.8\times {10}^{3}N}{\left(400\times {10}^{6}N/m^2\right)}\\ or,A_{\min }=2.45\times {10}^{-5}{m}^2\end{array}$

Hence, the minimum cross-sectional area of the wire is $2.45\times {10}^{-5}{m}^2$.

Conclusion:

Thus, the minimum cross-sectional area of the wire is $2.45\times {10}^{-5}{m}^2$.