Question

The resultant of two forces is 250 N and the same are inclined at 30° and 45° with resultant one on either side calculate the magnitude of two forces​

Answer 1

Answer:

The two forces are;

1) Force 1 with magnitude of approximately 183.013 N, acting 30° to the left of the resultant force

2) Force 2 with magnitude of approximately 129.41 N acting at an inclination of 45° to the right of the resultant force

Explanation:

The given parameters are;

The (magnitude) of the resultant of two forces = 250 N

The angle of inclination of the two forces to the resultant = 30° and 45°

Let, F₁ and F₂ represent the two forces, we have;

F₁ is inclined 30° to the left of the resultant force and F₂ is inclined 45° to the right of the resultant force

The components of F₁ are $\underset{F_1}{\rightarrow}$ = -F₁ × sin(30°)·i + F₁ × cos(30°)·j

The components of F₂ are $\underset{F_2}{\rightarrow}$ = F₂ × sin(45°)·i + F₂ × cos(45°)·j

The sum of the forces = F₂ × sin(45°)·i + F₂ × cos(45°)·j + (-F₁ × sin(30°)·i + F₁ × cos(30°)·j) = 250·j

The resultant force, R = 250·j, which is in the y-direction, therefore, the component of the two forces in the x-direction cancel out

We have;

F₂ × sin(45°)·i = F₁ × sin(30°)·i

F₂ ·√2/2 = F₁/2

∴ F₁ = F₂ ·√2

∴ F₂ × cos(45°)·j  + F₁ × cos(30°)·j = 250·j

Which gives;

F₂ × cos(45°)·j  + F₂ ·√2 × cos(30°)·j = 250·j

F₂ × ((cos(45°) + √2 × cos(30°))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ = 250·j/(((√2)/2 × (1 + √3))·j) ≈ 129.41 N

F₂ ≈ 129.41 N

F₁  = √2 × F₂ = √2 × 129.41 N ≈ 183.013 N

F₁  ≈ 183.013 N

The two forces are;

A force with magnitude of approximately 183.013 N is inclined 30° to the left of the resultant force and a force with magnitude of approximately 129.41 N is inclined 45° to the right of the resultant force.