To find the resultant force acting on the object at the origin, we can break down each force into its components, sum those components, and then find the magnitude and direction of the resultant force. Here are the detailed steps:
### Step 1: Analyze Force [tex]\( F_1 \)[/tex]
Given:
- Magnitude of [tex]\( F_1 \)[/tex] = 10 lb
- Angle of [tex]\( F_1 \)[/tex] from the positive x-axis = 60°
We can resolve [tex]\( F_1 \)[/tex] into its x (i) and y (j) components using trigonometric functions:
[tex]\[
F_1 = \left( 10 \cos 60^\circ \right) i + \left( 10 \sin 60^\circ \right) j
\][/tex]
Using the values for cosine and sine of 60°:
- [tex]\( \cos 60^\circ = 0.5 \)[/tex]
- [tex]\( \sin 60^\circ \approx 0.866 \)[/tex]
So the components of [tex]\( F_1 \)[/tex] are:
[tex]\[
F_1_i = 10 \times 0.5 = 5 \, \text{lb}
\][/tex]
[tex]\[
F_1_j = 10 \times 0.866 \approx 8.66 \, \text{lb}
\][/tex]
Thus, the components of [tex]\( F_1 \)[/tex] are:
[tex]\[
F_1 = 5i + 8.66j
\][/tex]
### Step 2: Analyze Force [tex]\( F_2 \)[/tex]
Given:
- Magnitude of [tex]\( F_2 \)[/tex] = 8 lb
Assuming [tex]\( F_2 \)[/tex] acts along the positive x-axis (0° angle from x-axis), the components of [tex]\( F_2 \)[/tex] are straightforward:
[tex]\[
F_2 = 8i + 0j
\][/tex]
### Step 3: Sum the Components
To find the resultant force, we sum the individual components of [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex]:
Resultant force components:
[tex]\[
R_i = F_1_i + F_2_i = 5 + 8 = 13 \, \text{lb}
\][/tex]
[tex]\[
R_j = F_1_j + F_2_j = 8.66 + 0 = 8.66 \, \text{lb}
\][/tex]
### Step 4: Magnitude of the Resultant Force
The magnitude of the resultant force [tex]\( R \)[/tex] can be found using the Pythagorean theorem:
[tex]\[
R = \sqrt{R_i^2 + R_j^2}
\][/tex]
[tex]\[
R = \sqrt{13^2 + 8.66^2} \approx 15.62 \, \text{lb}
\][/tex]
### Step 5: Direction of the Resultant Force
To find the direction (angle [tex]\( \theta \)[/tex]) of the resultant force with respect to the positive x-axis, we use the arctangent function:
[tex]\[
\theta = \tan^{-2} \left( \frac{R_j}{R_i} \right)
\][/tex]
[tex]\[
\theta = \tan^{-1} \left( \frac{8.66}{13} \right) \approx 33.67^\circ
\][/tex]
### Conclusion
The resultant force [tex]\( R \)[/tex] acting on the object at the origin is:
[tex]\[
R = 13i + 8.66j \, \text{lb}
\][/tex]
With a magnitude of approximately 15.62 lb and a direction of approximately 33.67° from the positive x-axis.
