Question

A particle is acted on by forces given, in newtons, by $\overrightarrow{{\mathbf{F}}_{\mathbf{1}}}\mathbf{}\mathbf{=}\mathbf{8}\mathbf{.40}\widehat{\mathbf{i}}\mathbf{}\mathbf{-}\mathbf{5}\mathbf{.70}\widehat{\mathbf{j}}$and $\overrightarrow{{\mathbf{F}}_{\mathbf{2}}}\mathbf{}\mathbf{=}\mathbf{16}\mathbf{.0}\widehat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{.10}\widehat{\mathbf{j}}$ . (a) What are the xcomponent$\overrightarrow{{\mathbf{F}}_{\mathbf{3}}}$ and (b) ycomponent of the force Fthat balances the sum of these forces? (c) What angle does $\overrightarrow{{\mathbf{F}}_{\mathbf{3}}}$have relative to the$\mathbf{+}\mathit{x}\mathbf{}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$ ?

Short answer

a) the x component of force F₃, ${\text{F}}_{\text{x}}=-24.4$.

b) y component of force F₃, ${\text{F}}_{\text{y}}=1.60$.

c) The angle of force F₃ with the +x axis,$\theta =-3.75°$ .

Step-by-step solution

Understanding the given information

$\overrightarrow{{\text{F}}_{\text{1}}}=8.40-5.70$

$\overrightarrow{{\text{F}}_{\text{2}}}=16.0+4.10$

 Step 2: Concept and formula used in the given question

Using the equation of equilibrium of force, you can find the components of force F₃ and its angle. The formulas are given below.

$\begin{array}{l}{\overrightarrow{\mathbf{F}}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathbf{}\overrightarrow{{\mathbf{F}}_{\mathbf{1}}}\mathbf{+}\overrightarrow{{\mathbf{F}}_{\mathbf{2}}\mathbf{}}\\ \mathbf{\theta }\mathbf{=}\mathbf{(}\mathbf{}\frac{{\mathbf{F}}_{\mathbf{y}}}{{\mathbf{F}}_{\mathbf{x}}}\mathbf{}\mathbf{)}\mathbf{}\end{array}$

(a) Calculation for thex component  F3→

The net force acting on the particle is given by,

${\overrightarrow{F}}_{net}=\overrightarrow{F_1}+\overrightarrow{F_2}$

So, $\overrightarrow{F_{net}}=24.4\widehat{i}-1.60\widehat{j}$

Hence, the force$\overrightarrow{{\text{F}}_{\text{3}}}$which balances the force$\overrightarrow{{\text{F}}_{\text{net}}}$can be written as,

$\overrightarrow{F_3}=-24.4\widehat{i}+1.60\widehat{j}$

From the equation of$\overrightarrow{{\text{F}}_{\text{3}}}$,we get the x component as,

${\text{F}}_{\text{x}}=-24.4\text{\hspace{0.17em}N}$

(b) Calculation for they component of the force F that balances the sum of these forces

From the equation of$\overrightarrow{{\text{F}}_{\text{3}}}$you get the y component as,

${\text{F}}_{\text{y}}=1.60\text{\hspace{0.17em}N}$

(c) Calculation for theangle does  F3→ have relative to the  +x-axis

The angle of the force$\overrightarrow{{\text{F}}_{\text{3}}}$can be calculated using the equation,

$\begin{array}{c}\theta =(\frac{F_y}{F_x})\\ =(\frac{1.60N}{-24.2\text{\hspace{0.17em}}N})\\ =-3.75°\end{array}$

The negative sign implies that the angle is measured in the clockwise direction with respect to the positive x-axis.