Physics, asked by abusiddiqui007, 4 months ago

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

Answered by Cosmique
24

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 ).


Anonymous: Nice
Answered by freedarajesh2003
2

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 ).

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