Question

4.6. Two forces, F, and F2, act at a point. The magnitude of #
is 9.00 N, and its direction is 60.0° above the x-axis in the sec.
ond quadrant. The magnitude of F2 is 6.00 N, and its direction is
53.1° below the x-axis in the third quadrant. (a) What are the x- and
y-components of the resultant force? (b) What is the magnitude of
the resultant force?​

Answer 1

Answer:

Question (a):

$\sum F_{x}=-8.1 \ N$

$\sum F_{y}=2.99 \ N$

Question (b):

$\left|F_{n e t}\right|=8.63 \ N$

Solution:

From question, the forces given are assumed as:

$F_{1}=9.00 N$

$F_{2}=6.00 \mathrm{N}$

Now, the x and y components of force $F_{1}$ are:

$F_{x 1}=-F_{1} \cos 60$ (Moves in negative x-direction)

$F_{y 1}=F_{2} \sin 60$ (Moves in positive y-direction)

Now, the x and y components of force $F_{2}$ are:

$F_{x 2}=-F_{2} \cos 53.1$ (Moves in negative x-direction)

$F_{y 2}=-F_{2} \sin 53.1$ (Moves in negative y-direction)

Question (a):

The x-component of resultant force:

$\sum F_{x}=F_{x 1}+F_{x 2}$

$\Rightarrow \sum F_{x}=-F_{1} \cos 60-F_{2} \cos 53.1$

On substituting the value of $F_1$,

$\Rightarrow \sum F_{x}=-9 \cos 60-6 \cos 53.1$

Thus, the resultant force of x components:

$\therefore \sum F_{x}=-8.1 \ N$

The y-component of resultant force:

$\sum F_{y}=F_{y 1}+F_{y 2}$

$\Rightarrow \sum F_{y}=F_{1} \sin 60-F_{2} \sin 53.1$

On substituting the value of $F_2$,

$\Rightarrow \sum F_{y}=9 \sin 60-6 \sin 53.1$

Thus, the resultant force of y components:

$\therefore \sum F_{y}=2.99 \ N$

Question (b):

The magnitude of resultant force is given by the formula:

$\left|F_{n e t}\right|=\sqrt{\left(\sum F_{x}\right)^{2}+\left(\sum F_{y}\right)^{2}}$

On substituting the values,

$\Rightarrow\left|F_{n e t}\right|=\sqrt{(-8.1)^{2}+(2.99)^{2}}$

$\Rightarrow\left|F_{n e t}\right|=\sqrt{65.61+8.94}$

$\Rightarrow\left|F_{n e t}\right|=\sqrt{74.55}$

$\therefore\left|F_{n e t}\right|=8.63 \ N$