Find the magnitude of the resultant force and the angle it makes with the positive $x$-axis. (Round your answers to one decimal place.)

• One force is $26 \, \text{lb}$ at $45^\circ$ above the $x$-axis,
• The other force is $16 \, \text{lb}$ at $30^\circ$ below the $x$-axis.

Neetesh Kumar | December 18, 2024

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This is the solution to Math 1C
Assignment: 12.2 Question Number 13
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Step-by-step solution:

We will resolve each force into its horizontal ($x$) and vertical ($y$) components, sum the components, and then calculate the magnitude and angle of the resultant force.

Step 1: Resolve forces into components:

1. Force 1: $F_1 = 26 \, \text{lb}$ at $45^\circ$ above the $x$-axis

   • Horizontal component: $F_{1x} = F_1 \cos(45^\circ)$
   • Vertical component: $F_{1y} = F_1 \sin(45^\circ)$

   Since $\cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071$:
   $F_{1x} = 26 \cdot 0.7071 = 18.4$,
   $F_{1y} = 26 \cdot 0.7071 = 18.4$.

2. Force 2: $F_2 = 16 \, \text{lb}$ at $30^\circ$ below the $x$-axis

   • Horizontal component: $F_{2x} = F_2 \cos(30^\circ)$
   • Vertical component: $F_{2y} = -F_2 \sin(30^\circ)$ (negative because it’s below the $x$-axis).

   Since $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660$ and $\sin(30^\circ) = 0.5$:
   $F_{2x} = 16 \cdot 0.86602540378 = 13.9$,
   $F_{2y} = -16 \cdot 0.5 = -8.0$.

Step 2: Find the resultant components:

The resultant force components are:
$F_x = F_{1x} + F_{2x}, \quad F_y = F_{1y} + F_{2y}.$

Substitute the values:
$F_x = 18.4 + 13.9 = 32.3$,
$F_y = 18.4 + (-8.0) = 10.4$.

Step 3: Find the magnitude of the resultant force:

The magnitude of the resultant force is given by:
$F = \sqrt{F_x^2 + F_y^2}.$

Substitute $F_x = 32.3$ and $F_y = 10.4$:
$F = \sqrt{(32.3)^2 + (10.4)^2}.$

Simplify:
$F = \sqrt{1043.3 + 108.2} = \sqrt{1151.5}.$

$F \approx 33.9 \, \text{lb}.$

Step 4: Find the angle with the positive $x$-axis:

The angle $\theta$ is given by:
$\theta = \arctan\left(\frac{F_y}{F_x}\right).$

Substitute $F_x = 32.3$ and $F_y = 10.4$:
$\theta = \arctan\left(\frac{10.4}{32.3}\right).$

Simplify:
$\theta = \arctan(0.322).$

Using a calculator:
$\theta \approx 17.8^\circ.$

Final Answers:

• Magnitude: $\boxed{33.9 \, \text{lb}}$

• Angle: $\boxed{17.9^\circ}$

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