91Ó°ÊÓ

The greatest extent of size and direction of an item is defined as magnitude.

Plastic ball mass =$1.0g$, string's length =$60cm$and the string's angle with the vertical = $20°$.

The arrangement's free body diagram is illustrated below.

The force equation in the vertical direction at equilibrium is,

$$\begin{gathered} T\cos \mathrm{θ}-mg=0 \\ T=\frac{mg}{\cos \mathrm{θ}} \end{gathered}$$

$T$= tension of string,

$m$= plastic ball mass,

$\mathrm{θ}$= string's angle with the vertical,

$g$= acceleration due to gravity.

The force equation in the horizontal direction at equilibrium is,

$$\begin{gathered} F_{elec}-T\sin \mathrm{θ}=0 \\ {F}_{elec}=T\sin \mathrm{θ} \end{gathered}$$

$F_{elec}$= magnitude of electric force.

Substitute $\frac{mg}{\cos \mathrm{θ}}$for $T$.

$$\begin{gathered} F_{elec}=\frac{mg}{\cos \mathrm{θ}}\sin \mathrm{θ} \\ =mg\tan \mathrm{θ} \end{gathered}$$

Substitute $1.0g$for $m$, $9.81m/s^2$for $g$and $20°$for $\mathrm{θ}$to find $\stackrel{→}{F_{elec}}$.

$$\begin{gathered} F_{elec}=\left(1.0g×\frac{{10}^{-3}kg}{1g}\right)\left(9.81m/s^2\right)\tan 20° \\ =0.0035kg·m/s^2×\frac{1N}{1kg·m/s^2} \\ =0.0035N \end{gathered}$$

Hence,$F_{elec}$magnitude is$0.0035N$.

The pulling force transferred axially by a string, cable, chain, or similar item, or by each end of a rod, truss member, or similar three-dimensional object is known as tension.

Plastic ball mass =$1.0g$, string's length =$60cm$and the string's angle with the vertical = $20°$.

Substitute $1.0g$for $m$, $9.81m/s^2$for $g$and $20°$for $\mathrm{θ}$in equation to find T.

$$\begin{gathered} T=\frac{mg}{\cos \mathrm{θ}} \\ T=\frac{\left(1.0g×{\displaystyle \frac{{10}^{-3}kg}{1g}}\right)\left(9.81m/s^2\right)}{\cos 20°} \\ =\frac{\left(0.00981kg·m/s^2\right)}{0.939} \\ =0.010kg·m/s^2×\frac{1N}{1kg·m/s^2} \\ =0.010N \end{gathered}$$

Hence, string's tension is$0.010N$.