Question

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


​

Answer 1

✯ 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,

magnitude of resultant force

• → R = √(3.35² + 3.67²)
• → R = 4.969 ≈ 5N.

✯ 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 2

Answer:

= (-4)5/(-4)8

= (-4)5-8

= 1/(-4)3