Question

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at an angle of ? = 65.0 ? above the negative xaxis in the second quadrant. F? 2 has a magnitudeof 5.80 N and is directed at an angle of ? = 53.9 ? below the negative x axis in the third quadrant.

Part A

What is the x component Fx of the resultant force?

Express your answer in newtons.

Part B

What is the y component Fy of the resultant force?

Express your answer in newtons.

Part C

What is the magnitude F of the resultant force?

Express your answer in newtons.

Part D

What is the angle ? that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.

Express your answer in degrees.

Two Forces Acting at a Point Two forces, F1 and F2, act at a point, as shown in the picture. (Fiqure 1) F1 has a magnitude of 9.20 N and is directed at an angle of α-65.0° above the negative x axis in the second quadrant. F2 has a magnitude of 5.80 N and is directed at an angle of β = 53.9" below the negative x axis in the third quadrant. Figure 1 of 1

Answer 1

Concepts and reason

The concept required to solve the problem are resolution of vectors and calculation of resultant vector. We have been given two force vectors with their angle with x axis. The two vectors can be resolved into their resultant components. Further, addition of their components along the respective axis will give the resultant vector’s components.

Fundamentals

The vector can be resolved into its components along the x and y axis as follow

$\begin{array}{l}\\{{\vec F}_{\rm{X}}} = |\vec F|\cos \theta \\\\{{\vec F}_{\rm{Y}}} = |\vec F|\sin \theta \\\end{array}$

The resultant vector can be given by

$\vec F = {\vec F_{\rm{X}}} + {\vec F_{\rm{Y}}}$

$\vec F = |\vec F|\cos \theta \hat i + |\vec F|\sin \theta \hat j$

Here, $\left| {\vec F} \right|$ is the magnitude of force vector and $\theta$ is the angle made by the vector with $x$ axis.

The magnitude of the resultant vector from the component is vector is given by

$F = \sqrt {{F_{\rm{X}}}^2 + {F_{\rm{Y}}}^2}$

The angle made by the vector with the x axis is given by

$\begin{array}{c}\\\tan \theta = \frac{{{F_{\rm{Y}}}}}{{{F_{\rm{X}}}}}\\\\\theta = {\tan ^{ - 1}}\frac{{{F_{\rm{Y}}}}}{{{F_{\rm{X}}}}}\\\end{array}$

(A)

The component of ${\vec F_1}$ in x direction is

${\vec F_{{\rm{x}}1}} = {\vec F_1}\cos {65^0}\left( { - \hat i} \right)$

Substitute 9.20 N for ${F_1}$

$\begin{array}{l}\\{{\vec F}_{{\rm{x}}1}} = \left( {9.20{\rm{N}}} \right)\cos {65^0}\left( { - \hat i} \right)\\\\{{\vec F}_{{\rm{x}}1}} = 3.88{\rm{N}}\left( { - \hat i} \right)\\\end{array}$

The component of ${\vec F_2}$ in x direction is

${\vec F_{{\rm{x}}2}} = {\vec F_2}\cos {53.9^0}\left( { - \hat i} \right)$

Substitute 5.80 N for ${F_2}$

$\begin{array}{l}\\{{\vec F}_{{\rm{x}}2}} = \left( {{\rm{5}}{\rm{.80N}}} \right)\cos {53.9^0}\left( { - \hat i} \right)\\\\{{\vec F}_{{\rm{x}}2}} = 3.417{\rm{N}}\left( { - \hat i} \right)\\\end{array}$

Therefore, the x component of resultant force is

${\vec F_{\rm{X}}} = {\vec F_{1{\rm{x}}}} + {\vec F_{2{\rm{x}}}}$

Substitute 3.88 N for ${F_{1{\rm{x}}}}$ and 3.417 for ${F_{2{\rm{x}}}}$

$\begin{array}{l}\\{{\vec F}_{\rm{X}}} = 3.88{\rm{N}}\left( { - \hat i} \right) + 3.417{\rm{N}}\left( { - \hat i} \right)\\\\{{\vec F}_{\rm{X}}} = 7.297\left( { - \hat i} \right)\\\end{array}$

(B)

The component of ${\vec F_1}$ in y direction is

${\vec F_{{\rm{y}}1}} = {\vec F_1}\sin {65^0}\left( {\hat j} \right)$

Substitute 9.20 N for ${F_1}$

$\begin{array}{l}\\{{\vec F}_{{\rm{y}}1}} = \left( {9.20{\rm{N}}} \right)\sin {65^0}\left( {\hat j} \right)\\\\{{\vec F}_{{\rm{y}}1}} = 8.33{\rm{N}}\left( {\hat j} \right)\\\end{array}$

The component of ${\vec F_2}$ in y direction is

${\vec F_{{\rm{y}}2}} = {\vec F_2}\sin {53.9^0}\left( { - \hat j} \right)$

Substitute 5.80 N for ${F_2}$

$\begin{array}{l}\\{{\vec F}_{{\rm{y}}2}} = \left( {{\rm{5}}{\rm{.80N}}} \right)\sin {53.9^0}\left( { - \hat j} \right)\\\\{{\vec F}_{{\rm{y}}2}} = 4.68{\rm{N}}\left( { - \hat j} \right)\\\end{array}$

Therefore, the y component of resultant force is

$\begin{array}{l}\\{{\vec F}_{\rm{Y}}} = {{\vec F}_{{\rm{1y}}}} + {{\vec F}_{{\rm{2y}}}}\\\\{{\vec F}_{\rm{Y}}} = 8.33{\rm{N}}\left( {\hat j} \right) + 4.68{\rm{N}}\left( { - \hat j} \right)\\\\{{\vec F}_{\rm{Y}}} = 3.65{\rm{N}}\left( {\hat j} \right)\\\end{array}$

(C)

The resultant vector is given by

$\vec F = {\vec F_{\rm{X}}} + {\vec F_{\rm{Y}}}$

Substitute 7.297 N for ${F_{\rm{X}}}$ and 3.65 N for ${F_{\rm{Y}}}$

$\vec F = 7.297\left( { - \hat i} \right) + 3.65{\rm{N}}\left( {\hat j} \right)$

Therefore, the magnitude of Resultant vector is

$\left| F \right| = \sqrt {{F_{\rm{X}}}^2 + {F_{\rm{Y}}}^2}$

Substitute 7.297 N for ${F_{\rm{X}}}$ and 3.65 N for ${F_{\rm{Y}}}$

$\begin{array}{l}\\\left| F \right| = \sqrt {{{\left( {7.297{\rm{N}}} \right)}^2} + {{\left( {3.65{\rm{N}}} \right)}^2}} \\\\\left| F \right| = \sqrt {53.24 + 13.32} \\\\\left| F \right| = 8.158{\rm{N}}\\\end{array}$

(D)

The resultant vector is

$\vec F = 7.297\left( { - \hat i} \right) + 3.65{\rm{N}}\left( {\hat j} \right)$

Therefore, the angle made with -x axis

$\tan \theta = \frac{{{F_{\rm{Y}}}}}{{{F_{\rm{X}}}}}$

Substitute 7.297 N for ${F_{\rm{X}}}$ and 3.65 N for ${F_{\rm{Y}}}$

$\begin{array}{l}\\\tan \theta = \frac{{3.65}}{{7.297}}\\\\\tan \theta = 0.50\\\\\theta = {\tan ^{ - 1}}\left( {0.50} \right)\\\\\theta = {26.56^0}\\\end{array}$

It is positive since the resultant vector is in the second quadrant and the angle measured in clockwise direction from -x is taken positive as per the question.

Ans: Part A

The x component of resultant force is ${\vec F_{\rm{X}}} = 7.297\left( { - \hat i} \right)$

Part B

The y component of resultant force is ${\vec F_{\rm{Y}}} = 3.65{\rm{N}}\left( {\hat j} \right)$

Part C

The magnitude of the resultant vector is $\left| F \right| = 8.158{\rm{N}}$

Part D

The angle of inclination with the -x axis is $\theta = {26.56^0}$