Chapter 5.3, Problem 43SSC

To determine

The mass of the scoreboard.

Answer to Problem 43SSC

The mass of the scoreboard is $m=1311\text{kg}$ .

Explanation of Solution

Given:

Consider a scoreboard is suspended from the top of the ceiling with the help of ten cables.

The angle made by the 6 cables with the vertical is $\alpha =8.0°$ .

The angle made by the remaining 4 cables with the vertical is $\beta =10.0°$ .

The tension in each of the 10 cables is $T=1300\text{N}$ .

Consider the mass of the scoreboard is $m$ .

Formula used:

Consider a cable is inclined at an $\theta$ with the vertical, and connects the ceiling with the scoreboard as shown in figure below.

Consider a force $F$ is acting in a cable inclined at angle $\theta$ with the vertical.

The horizontal component of the force $F$ is,

${\left(F\right)}_{\text{H}}=F\sin \theta$

The vertical component of the force $F$ is,

${\left(F\right)}_{\text{V}}=F\cos \theta$

Calculation:

Consider 10 cables are used to connect the ceiling with the scoreboard.

The tension in each cable is $T=1300\text{N}$ .

Six cables are inclined at angle $\alpha =8.0°$ with the vertical.

The combined vertical component of the tension $T$ in the six cables is,

$\begin{array}{c}{F}_{\left(\begin{array}{l}\text{Vertical component}\\ \text{ of 6 cables}\end{array}\right)}=6\times T\cos \alpha \\ =6\times 1300\cos \left(8.0°\right)\\ =7724\text{N}\end{array}$

Remaining four cables are inclined at angle $\beta =10.0°$ with the vertical.

The combined vertical component of the tension $T$ in the four cables is,

$\begin{array}{c}{F}_{\left(\begin{array}{l}\text{Vertical component}\\ \text{ of 4 cables}\end{array}\right)}=4\times T\cos \beta \\ =4\times 1300\cos \left(10.0°\right)\\ =5121\text{N}\end{array}$

The net vertical component of the tension in the 10 cables is,

$\begin{array}{c}{F}_{\left(\begin{array}{l}\text{Vertical component}\\ \text{ of 10 cables}\end{array}\right)}=F_{\left(\begin{array}{l}\text{Vertical component}\\ \text{ of 6 cables}\end{array}\right)}+F_{\left(\begin{array}{l}\text{Vertical component}\\ \text{ of 4 cables}\end{array}\right)}\\ =7724\text{N}+5121\text{N}\\ =\text{12845}\text{N}\end{array}$

Consider the value of acceleration due to gravity as $g=9.80{\text{m/s}}^2$ .

The gravitational force acting on the scoreboard is,

$\begin{array}{c}{F}_{\text{g}}=mg\\ =m\times 9.80\end{array}$

The net vertical component of the tension in the 10 cables balances the gravitational force of the scoreboard.

$\begin{array}{l}{F}_{\left(\begin{array}{l}\text{Vertical component}\\ \text{ of 10 cables}\end{array}\right)}=F_{\text{g}}\\ 12845\text{N}\times \left(\frac{1\text{kg}\cdot {\text{m/s}}^2}{1\text{N}}\right)=m\times \left(9.80{\text{m/s}}^2\right)\\ m=\left(\frac{12845}{9.80}\right)\text{kg}\\ m=1311\text{kg}\end{array}$

Conclusion:

Thus, the mass of the scoreboard is $m=1311\text{kg}$ .