Four forces of magnitude 2N, 3N,4N and 5N are acting at a point at angles of 30°,45°,60° and
120° respectively with X-axis. If the forces are coplanar then calculate the magnitude and
direction of their resultant
Answers
Given:
- Four forces of magnitudes 2N, 3N, 4N and 5N are acting at angles of 30°, 45°, 60° and 120° respectively with the x-axis.
- Forces are co-planar.
To find:
- Magnitude and direction of resultant
Knowledge required:
Equilibrium of coplanar forces
- When more than a single force is acting on a body, then they can be resolved into its rectangular components. And, the sum of all x-components will give us the x-component of the resultant force, and the sum of all y-components will give us the y-component of the resultant force.
- Resolution of vectors
For a vector 'V' making an angle 'a' with the x-axis in an x-y plane., it can be resolved into its corresponding vectors along the x-axis and the y-axis as V cos a and V sin a respectively.
Solution:
According to the given data and using the concept of resolution of vectors,
The x-component of
2N force will be 2 cos 30°
3N force will be 3 cos 45°
4N force will be 4 cos 60°
5 N force will be 5 cos 120°
So,
The x-component of the resultant vector will be,
→ R_x = 2 cos 30° + 3 cos 45° + 4 cos 60° + 5 cos 120°
→ R_x = 2 · √3/2 + 3 · 1/√2 + 4 · 1/2 + 5 · -1/2
→ R_x = ( 2√3 + 3√2 -1 ) / 2
→ R_x = ( 2 · 1.73 + 3 · 1.41 - 1 ) / 2 = 3.35 N
And, the y-component of
2N force will be 2 sin 30°
3N force will be 3 sin 45°
4N force will be 4 sin 60°
5N force will be 5 sin 120°
So, the y-component of the resultant vector will be,
→ R_y = 2 sin 30° + 3 sin 45° + 4 sin 60° + 5 sin 120°
→ R_y = 2 · 1/2 + 3 · 1/√2 + 4 · √3/2 + 5 · -1/2
→ R_y = ( 3√2 + 4√3 - 3 ) / 2
→ R_y = ( 3 · 1.14 + 4 · 1.73 - 3 ) / 2 = 3.67 N
Therefore,
Our resultant force will be,
→ R = R_x + R_y
→ R = 3.35 i + 3.67 j N
so,
the magnitude of resultant force
→ R = √(3.35² + 3.67²)
→ R = 4.969 ≈ 5N.
and, For the direction of R
Let resultant force is at an angle θ with the x-axis., then
→ tan θ = R_y / R_x
→ tan θ = 3.67 / 3.35
→ θ = tan⁻¹ ( 1.09 ).
Answer:
Explanation:
Given:
Four forces of magnitudes 2N, 3N, 4N and 5N are acting at angles of 30°, 45°, 60° and 120° respectively with the x-axis.
Forces are co-planar.
To find:
Magnitude and direction of resultant
Knowledge required:
Equilibrium of coplanar forces
When more than a single force is acting on a body, then they can be resolved into its rectangular components. And, the sum of all x-components will give us the x-component of the resultant force, and the sum of all y-components will give us the y-component of the resultant force.
Resolution of vectors
For a vector 'V' making an angle 'a' with the x-axis in an x-y plane., it can be resolved into its corresponding vectors along the x-axis and the y-axis as V cos a and V sin a respectively.
Solution:
According to the given data and using the concept of resolution of vectors,
The x-component of
2N force will be 2 cos 30°
3N force will be 3 cos 45°
4N force will be 4 cos 60°
5 N force will be 5 cos 120°
So,
The x-component of the resultant vector will be,
→ R_x = 2 cos 30° + 3 cos 45° + 4 cos 60° + 5 cos 120°
→ R_x = 2 · √3/2 + 3 · 1/√2 + 4 · 1/2 + 5 · -1/2
→ R_x = ( 2√3 + 3√2 -1 ) / 2
→ R_x = ( 2 · 1.73 + 3 · 1.41 - 1 ) / 2 = 3.35 N
And, the y-component of
2N force will be 2 sin 30°
3N force will be 3 sin 45°
4N force will be 4 sin 60°
5N force will be 5 sin 120°
So, the y-component of the resultant vector will be,
→ R_y = 2 sin 30° + 3 sin 45° + 4 sin 60° + 5 sin 120°
→ R_y = 2 · 1/2 + 3 · 1/√2 + 4 · √3/2 + 5 · -1/2
→ R_y = ( 3√2 + 4√3 - 3 ) / 2
→ R_y = ( 3 · 1.14 + 4 · 1.73 - 3 ) / 2 = 3.67 N
Therefore,
Our resultant force will be,
→ R = R_x + R_y
→ R = 3.35 i + 3.67 j N
so,
the magnitude of resultant force
→ R = √(3.35² + 3.67²)
→ R = 4.969 ≈ 5N.
and, For the direction of R
Let resultant force is at an angle θ with the x-axis., then
→ tan θ = R_y / R_x
→ tan θ = 3.67 / 3.35
→ θ = tan⁻¹ ( 1.09 ).