Two forces 8N & 10N inclined at an angle of 30¡ã are acting on a particle. find the resultant force and its direction.
you can consider one of the forces along the x-axis. Let's use the 10N one. Then the two forces are
<10,0> and <6.93,4.0>
Add hem up to get
<16.93,4.0>
or
17.40N at an angle of 13.3° from the 10N force.
IS THAT RIGHT?[14.546]
To find the resultant force and its direction, you can use vector addition.
1. Draw a diagram: Draw a coordinate system and place the two forces as vectors.
- The vector of 8N should be inclined at an angle of 30°.
- The vector of 10N should be inclined at an angle of 0° (since it is horizontal).
2. Resolve the forces: Resolve each force into horizontal and vertical components.
- The horizontal component of the 8N force can be found by multiplying it by cos(30°).
- The vertical component of the 8N force can be found by multiplying it by sin(30°).
- The horizontal component of the 10N force is equal to the force itself, as it is already horizontal.
3. Determine the values of the components:
- The horizontal component of the 8N force: 8N * cos(30°) = 6.93 N (approximately)
- The vertical component of the 8N force: 8N * sin(30°) = 4 N (approximately)
- The horizontal component of the 10N force: 10N
4. Add the horizontal components together:
- 6.93 N + 10 N = 16.93 N (approximately)
5. Add the vertical components together:
- 4 N
6. Find the magnitude of the resultant force:
- Use the Pythagorean theorem: Resultant Force = √(16.93 N^2 + 4 N^2)
- Resultant Force ≈ √289.53 N^2 ≈ 17 N (approximately)
7. Find the direction of the resultant force:
- The angle can be found using the inverse tangent function (tan^-1) of the vertical component divided by the horizontal component.
- Resultant Force's angle = tan^-1(4 N / 16.93 N) ≈ 13.8° (approximately)
(Note: The tangent of an angle is the opposite/adjacent sides, i.e., vertical/horizontal components in this case.)
Therefore, the resultant force is approximately 17 N and its direction is approximately 13.8° above the horizontal.