Question

C) Five forces of 100, 200, 300, 400 and 500 N are acting at angles of 45, 100, 210, 280 and 340 degrees in anti-clockwise direction from x-axis, at a point all acting away from the point. Find the resultant force in magnitude and direction.​

Answer 1

Explanation:

A = 100 N

B = 150 N

Angle between A & B, θ = 45°

Resultant Force, |C| = √[|A|^2 +|B|^2 + 2|A||B|cosθ]

=> |C| = 231.73 N at an angle of ϕ = 27.23° from the direction of 100 N force.

tan ϕ = bsinθ/(a + bcosθ)

Answer 2

Answer:

Magnitude of Resultant Force is 565.153 Newton

Direction of this force is θ = - 54.8° with positive X axis (Clockwise direction from X-axis)

Explanation:

Given that:

Five forces of 100, 200, 300, 400 and 500 N are acting at angles of 45, 100, 210, 280 and 340 degrees in anti-clockwise direction from x-axis

So, Let:

The forces are f1, f2, f3, f4, f5 and their components towards positive X axis are-

f1 Cosθ1, f2 Cosθ2, f3 Cosθ3, f4 Cosθ4, f5 Cosθ5 respectively;

So we get:

$f1 = 100 \\ f1 \cos(45) = 70.71 \\ \\ f2 = 200 \\ f2 \cos(100) = - 34.73$

$f3 = 300 \\ f3 \cos(210) = - 259.81 \\ \\ f4 = 400 \\ f4 \cos(280) = 69.46$

$f5 = 500 \\ f5 \cos(340) = 469.85$

Sum of these components = Resultant Force towards X axis ;

Fx = f1 Cosθ1 + f2 Cosθ2 + f3 Cosθ3 + f4 Cosθ4 + f5Cosθ5

= 315.48 N

Similarly, their components towards positive Y axis are-

f1 Sinθ1, f2 Sinθ2, f3 Sin θ3, f4 Sinθ4, f5 Sin θ5 respectively;

So we get:

$f1 = 100 \\ f1 \ \sin(45) = 70.71 \\ \\ f2 = 200 \\ f2 \ \sin(100) = 196.96$

$f3 = 300 \\ f3 \sin(210) = - 150 \\ \\ f4 = 400 \\ f4 \sin(280) = - 393.92$

$f5 = 500 \\ f5 \sin(340) = - 171.01$

Sum of these components = Resultant Force towards Y axis;

Fy = f1 Sinθ1 + f2 Sinθ2 + f3 Sin θ3 +f4 Sinθ4 + f5 Sinθ5

=-447.26 N

So, Resultant Force = Fx + Fy

Fr = √[(Fx)^2 + (Fy)^2] = 565.153 N

Direction of this force :

θ = tan ^-1(Fy / Fx) = tan ^-1( -447.26 / 315.48 )

= -54.8° with positive X axis

So the resultant force is acting at 54.8° clockwise direction from X-axis.