Two forces [tex]$30 \, \text{N}$[/tex] and [tex]$40 \, \text{N}$[/tex] act at right angles to each other. Determine, by scale drawing, the magnitude and direction of the resultant force, using a scale of 1 cm to 10 N.



Answer :

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].