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To determine the magnitude and direction of the resultant force when two forces act at right angles to each other, you can follow these steps:
### Step 1: Identifying the forces
We have two forces:
- Force 1 ([tex]\( F_1 \)[/tex]) = 30 units
- Force 2 ([tex]\( F_2 \)[/tex]) = 40 units
These forces are acting perpendicular to each other.
### Step 2: Represent the forces on a coordinate system
Since these forces are at right angles, you can place them on a coordinate system:
- Let [tex]\( F_1 \)[/tex] act along the x-axis.
- Let [tex]\( F_2 \)[/tex] act along the y-axis.
On your coordinate system, you would draw:
- A horizontal line, 30 cm long, to represent [tex]\( F_1 \)[/tex].
- A vertical line, 40 cm long, to represent [tex]\( F_2 \)[/tex].
### Step 3: Draw the resultant force
To find the resultant force:
- Draw a vector from the origin to the point where [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex] meet (forming a right-angled triangle).
### Step 4: Calculate the magnitude of the resultant force
The magnitude of the resultant force ([tex]\( R \)[/tex]) can be determined using the Pythagorean theorem:
[tex]\[ R = \sqrt{F_1^2 + F_2^2} \][/tex]
Given that [tex]\( F_1 = 30 \)[/tex] units and [tex]\( F_2 = 40 \)[/tex] units:
[tex]\[ R = \sqrt{30^2 + 40^2} \][/tex]
[tex]\[ R = \sqrt{900 + 1600} \][/tex]
[tex]\[ R = \sqrt{2500} \][/tex]
[tex]\[ R = 50 \][/tex]
So, the magnitude of the resultant force is 50 units.
### Step 5: Determine the direction of the resultant force
To find the direction of the resultant force, we need to calculate the angle ([tex]\( \theta \)[/tex]) it makes with the x-axis (horizontal line). This can be done using the tangent function:
[tex]\[ \tan(\theta) = \frac{F_2}{F_1} \][/tex]
Given that [tex]\( F_2 = 40 \)[/tex] units and [tex]\( F_1 = 30 \)[/tex] units:
[tex]\[ \tan(\theta) = \frac{40}{30} \][/tex]
[tex]\[ \theta = \tan^{-1}\left(\frac{40}{30}\right) \][/tex]
Calculating the inverse tangent:
[tex]\[ \theta \approx 53.13^\circ \][/tex]
### Conclusion
The magnitude of the resultant force is 50 units, and the direction (angle) with respect to the x-axis is approximately [tex]\( 53.13^\circ \)[/tex].
### Step 1: Identifying the forces
We have two forces:
- Force 1 ([tex]\( F_1 \)[/tex]) = 30 units
- Force 2 ([tex]\( F_2 \)[/tex]) = 40 units
These forces are acting perpendicular to each other.
### Step 2: Represent the forces on a coordinate system
Since these forces are at right angles, you can place them on a coordinate system:
- Let [tex]\( F_1 \)[/tex] act along the x-axis.
- Let [tex]\( F_2 \)[/tex] act along the y-axis.
On your coordinate system, you would draw:
- A horizontal line, 30 cm long, to represent [tex]\( F_1 \)[/tex].
- A vertical line, 40 cm long, to represent [tex]\( F_2 \)[/tex].
### Step 3: Draw the resultant force
To find the resultant force:
- Draw a vector from the origin to the point where [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex] meet (forming a right-angled triangle).
### Step 4: Calculate the magnitude of the resultant force
The magnitude of the resultant force ([tex]\( R \)[/tex]) can be determined using the Pythagorean theorem:
[tex]\[ R = \sqrt{F_1^2 + F_2^2} \][/tex]
Given that [tex]\( F_1 = 30 \)[/tex] units and [tex]\( F_2 = 40 \)[/tex] units:
[tex]\[ R = \sqrt{30^2 + 40^2} \][/tex]
[tex]\[ R = \sqrt{900 + 1600} \][/tex]
[tex]\[ R = \sqrt{2500} \][/tex]
[tex]\[ R = 50 \][/tex]
So, the magnitude of the resultant force is 50 units.
### Step 5: Determine the direction of the resultant force
To find the direction of the resultant force, we need to calculate the angle ([tex]\( \theta \)[/tex]) it makes with the x-axis (horizontal line). This can be done using the tangent function:
[tex]\[ \tan(\theta) = \frac{F_2}{F_1} \][/tex]
Given that [tex]\( F_2 = 40 \)[/tex] units and [tex]\( F_1 = 30 \)[/tex] units:
[tex]\[ \tan(\theta) = \frac{40}{30} \][/tex]
[tex]\[ \theta = \tan^{-1}\left(\frac{40}{30}\right) \][/tex]
Calculating the inverse tangent:
[tex]\[ \theta \approx 53.13^\circ \][/tex]
### Conclusion
The magnitude of the resultant force is 50 units, and the direction (angle) with respect to the x-axis is approximately [tex]\( 53.13^\circ \)[/tex].
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