Question Find the magnitude of a force pulling $45.8^{\circ}$ west and the direction of the $166.2 \mathrm{N}$ resultant force when a $116 \mathrm{N}$ force pulls due west.

Studdy Solution

STEP 1

Assumptions
1. The first force, $F_1$ , has a magnitude of $116 \mathrm{~N}$ and is directed due west.
2. The second force, $F_2$ , has an unknown magnitude and is directed $45.8^{\circ}$ west of north.
3. The resultant force, $F_R$ , has a magnitude of $166.2 \mathrm{~N}$ .
4. The direction of the resultant force is also unknown and is to be expressed as $\mathrm{N}\theta^{\circ}\mathrm{W}$ , where $\theta$ is the angle to be found.
5. Forces are vectors and can be added using vector addition.
6. The problem is in two dimensions and can be solved using trigonometry and vector components.

STEP 2

Decompose the first force, $F_1$ , into its vector components. Since $F_1$ is directed due west, it only has a component in the westward direction (negative x-axis).
$F_{1x} = -116 \mathrm{~N}$ $F_{1y} = 0 \mathrm{~N}$

STEP 3

Decompose the second force, $F_2$ , into its vector components. Since $F_2$ is directed $45.8^{\circ}$ west of north, it has components in both the northward (positive y-axis) and westward (negative x-axis) directions.
$F_{2x} = -|F_2| \cos(45.8^{\circ})$ $F_{2y} = |F_2| \sin(45.8^{\circ})$

STEP 4

The resultant force, $F_R$ , is the vector sum of $F_1$ and $F_2$ .
$F_{Rx} = F_{1x} + F_{2x}$ $F_{Ry} = F_{1y} + F_{2y}$

STEP 5

Given the magnitude of the resultant force, we can express it in terms of its components.
$|F_R| = \sqrt{F_{Rx}^2 + F_{Ry}^2}$

STEP 6

Substitute the known values and expressions for the components of $F_R$ into the equation for its magnitude.
$166.2 \mathrm{~N} = \sqrt{(-116 \mathrm{~N} - |F_2| \cos(45.8^{\circ}))^2 + (|F_2| \sin(45.8^{\circ}))^2}$

STEP 7

Square both sides of the equation to eliminate the square root.
$(166.2 \mathrm{~N})^2 = (-116 \mathrm{~N} - |F_2| \cos(45.8^{\circ}))^2 + (|F_2| \sin(45.8^{\circ}))^2$

STEP 8

Expand the squares on the right side of the equation.
$27624.84 \mathrm{~N}^2 = (116 \mathrm{~N})^2 + 2 \cdot 116 \mathrm{~N} \cdot |F_2| \cos(45.8^{\circ}) + (|F_2| \cos(45.8^{\circ}))^2 + (|F_2| \sin(45.8^{\circ}))^2$

STEP 9

Combine like terms and simplify the equation.
$27624.84 \mathrm{~N}^2 = 13456 \mathrm{~N}^2 + 2 \cdot 116 \mathrm{~N} \cdot |F_2| \cos(45.8^{\circ}) + |F_2|^2 (\cos^2(45.8^{\circ}) + \sin^2(45.8^{\circ}))$

STEP 10

Use the Pythagorean identity $\cos^2(\theta) + \sin^2(\theta) = 1$ to simplify the equation further.
$27624.84 \mathrm{~N}^2 = 13456 \mathrm{~N}^2 + 2 \cdot 116 \mathrm{~N} \cdot |F_2| \cos(45.8^{\circ}) + |F_2|^2$

STEP 11

Subtract $13456 \mathrm{~N}^2$ from both sides of the equation.
$27624.84 \mathrm{~N}^2 - 13456 \mathrm{~N}^2 = 2 \cdot 116 \mathrm{~N} \cdot |F_2| \cos(45.8^{\circ}) + |F_2|^2$

STEP 12

Calculate the left side of the equation.
$14168.84 \mathrm{~N}^2 = 2 \cdot 116 \mathrm{~N} \cdot |F_2| \cos(45.8^{\circ}) + |F_2|^2$

STEP 13

To find $|F_2|$ , we need to solve the quadratic equation. Let's rearrange the terms to get a standard quadratic equation form $ax^2 + bx + c = 0$ .
$|F_2|^2 + 2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ}) \cdot |F_2| - 14168.84 \mathrm{~N}^2 = 0$

STEP 14

Now we can use the quadratic formula to solve for $|F_2|$ .
$|F_2| = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Where $a = 1$ , $b = 2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ})$ , and $c = -14168.84 \mathrm{~N}^2$ .

STEP 15

Calculate the value of $b$ .
$b = 2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ})$

STEP 16

Calculate the value of $b^2$ .
$b^2 = (2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ}))^2$

STEP 17

Substitute the values of $a$ , $b$ , and $c$ into the quadratic formula.
$|F_2| = \frac{-(2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ})) \pm \sqrt{(2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ}))^2 - 4 \cdot 1 \cdot (-14168.84 \mathrm{~N}^2)}}{2 \cdot 1}$

STEP 18

Simplify the expression under the square root (the discriminant).
$\Delta = (2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ}))^2 + 4 \cdot 14168.84 \mathrm{~N}^2$

STEP 19

Calculate the discriminant, $\Delta$ .
$\Delta = (2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ}))^2 + 4 \cdot 14168.84 \mathrm{~N}^2$

STEP 20

Substitute the value of the discriminant back into the quadratic formula.
$|F_2| = \frac{-(2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ})) \pm \sqrt{\Delta}}{2}$

STEP 21

Since we are looking for a magnitude, which must be positive, we will use the positive square root.
$|F_2| = \frac{-(2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ})) + \sqrt{\Delta}}{2}$

STEP 22

Calculate the magnitude of the second force, $|F_2|$ .
$|F_2| = \frac{-(2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ})) + \sqrt{\Delta}}{2}$

STEP 23

Now that we have the magnitude of the second force, we need to find the direction of the resultant force. The direction can be found by calculating the angle $\theta$ for the resultant force using its components.
$\tan(\theta) = \frac{F_{Ry}}{F_{Rx}}$

STEP 24

Substitute the expressions for $F_{Rx}$ and $F_{Ry}$ .
$\tan(\theta) = \frac{|F_2| \sin(45.8^{\circ})}{-116 \mathrm{~N} - |F_2| \cos(45.8^{\circ})}$

STEP 25

Calculate the angle $\theta$ using the arctangent function.
$\theta = \arctan\left(\frac{|F_2| \sin(45.8^{\circ})}{-116 \mathrm{~N} - |F_2| \cos(45.8^{\circ})}\right)$

STEP 26

The direction of the resultant force will be north of west, so we need to ensure that the angle $\theta$ is positive.
$\theta = \left|\arctan\left(\frac{|F_2| \sin(45.8^{\circ})}{-116 \mathrm{~N} - |F_2| \cos(45.8^{\circ})}\right)\right|$

STEP 27

Now we have all the information to write the final answer. Calculate the exact values of $|F_2|$ and $\theta$ using the equations derived in previous steps.
$\left|F_{2}\right|=\frac{-(2 \cdot 116 \mathrm{~N} \cdot \cos(45.8^{\circ})) + \sqrt{\Delta}}{2} \mathrm{~N}$
$\theta = \left|\arctan\left(\frac{|F_2| \sin(45.8^{\circ})}{-116 \mathrm{~N} - |F_2| \cos(45.8^{\circ})}\right)\right|^{\circ}$
The magnitude of the second force is $\left|F_{2}\right|$ N, and the direction of the resultant is $\mathrm{N}\theta^{\circ} \mathrm{W}$ .

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