Question

1. Two forces of magnitudes 50 kN and 80 kN are acting on a particle, such that the angle between the
two is 135 degree. If both the forces are acting away from the particle, calculate the resultant and
find its direction.​

Answer 1

Answer:

Magnitude F = √ (Fx 2 +Fy 2 )

                       =√ (7002 + 15002 )

                        = 1655.29N

Angle of he force with horizontal θx = cos-1 (Fx/F) θx

                                                           = cos-1 (700/1655.29)

                                                           = 64.98

Answer 2

Answer:

Two forces of magnitudes 50 kN and 80 kN are acting on a particle, such that the angle between the two is 135 degree. If both the forces are acting away from the particle, The magnitude of resultant is R= 56.95 kN

The angle of direction of resultant will be α = -83.38 °.

Explanation:

Let, F₁= 50 kN and F₂= 80 kN

θ= 135°

Due to the fact that the forces are acting on a particle, the given system of two forces is a system of coplanar concurrent forces.

By using the following analytical method to solve the problem it is solved.

The Resultant using the force parallelogram: In the step 1 Draw the force parallelogram, not necessarily to scale, and label it with all the known values.

These are the characteristics of the resulting force:

The magnitude of resultant is=

$R = \sqrt{ 50^{2} + 80^{2} + 2 (50)(80)cos135 }$

= 56.95 kN

The angle of resultant will be=

α = tan⁻¹ [ F₂ sinθ/ (F₁+ F₂Sinθ)]

= -83.38 °.

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