Find the magnitude of a fourth force on the stone that will make the vector sum of the four forces zero.

100cos30° + 80cos120° + 40cos233° + Dx = 0
Dx = -22.53 N

y components:
100sin30° + 80sin120° + 40sin233° + Dy = 0
Dy = -87.34 N

magnitude of D:
sqrt[(-22.53)² + (-87.34)²]
90.2 N

direction of D:
arctan[(-87.34)/(-22.53)]
75.5° ref, but since Dx and Dy are both negative, we know this vector is in QIV:
360 - 75.5° = 284.5°

Answer Link

ap8997154 ap8997154 · Answer 2 · 2022-02-02T12:53:14+01:00

The magnitude of the fourth force will be 90.1956 N and the angle will be 75.535° made from the horizontal in the first quadrant.

Given to us

Force A, [tex]F_A = 100\ N[/tex]
Angle made by Force A from the horizontal, [tex]\theta_A = 30^o[/tex]
Force B, [tex]F_B = 80 \ N[/tex]
Angle made by Force B from the Vertical, [tex]\theta_B = 30^o[/tex]
Force C, [tex]F_C = 40 \ N[/tex]
Angle made by Force C from the Horizontal, [tex]\theta_C = 53^o[/tex]

For Angle made by Force B,

As the angle given to us is from the vertical axis, we will make it from the horizontal,

[tex]\theta_B = (90^o -30^o) = 60^o[/tex]

Now, the magnitude of a fourth force on the stone should make the vector sum of the four forces zero. therefore,

[tex]\sum F_{horizontal} = 0[/tex]

[tex]F_A Cos(\theta_A) -F_B Cos(\theta_B) -F_C Cos(\theta_C) = F_{Dh}[/tex]

[tex]100 Cos(30^o) -80Cos(60^o) -40Cos(53^o) = F_{Dh}\\\\F_{Dh} = 22.5299\ N[/tex]

[tex]\sum F_{Vertical} = 0[/tex]

[tex]F_A Sin(\theta_A) +F_B Sin(\theta_B) -F_C Sin(\theta_C) = F_{Dv}[/tex]

[tex]100 Sin(30^o) +80Sin(60^o) -40Sin(53^o) = F_{Dv}\\\\F_{Dv} =87.3365\ N[/tex]

Resultant Force

[tex]R = \sqrt{(F_{Dh})^2+(F_{Dv})^2}\\\\R = \sqrt{(87.3365)^2+(22.5299)^2}\\\\R = \sqrt{7627.6642+507.5963}\\\\R = \sqrt{813526}\\\\R = 90.1956\ N[/tex]

The angle of the resultant,

[tex]Tan (\theta_D) = \dfrac{F_{Dv}}{F_{Dh}}\\\\Tan(\theta_D) = \dfrac{87.3365}{22.5299}\\\\Tan(\theta_D) = 3.87647\\\theta_D = Tan^{-1} 3.87647\\\theta_D =75.535^o[/tex]

Hence, the magnitude of the fourth force will be 90.1956 N and the angle will be 75.535° made from the horizontal in the first quadrant.

Learn more about Forces:

https://brainly.com/question/14866438