Respuesta :
Answer:
The magnitude of third force is 74.4 N and direction of third force is 72.8 degrees South.
Explanation:
Let F1, F2 and F3 are three forces exerted on an object.
[tex]\theta_1=90^{\circ}[/tex]
[tex]\theta_2=60^{\circ}[/tex]
[tex]|F_1|=33 N[/tex]
[tex]|F_2|=44 N[/tex]
We have to find the direction and magnitude of third force i.e F3.
[tex]F_{1x}=33cos(90^{\circ})=0 N[/tex]
[tex]F_{1y}=33sin(90^{\circ})=33 N[/tex]
[tex]F_{2x}=44cos(60^{\circ})=22 N[/tex]
[tex]F_{2y}=44 sin(60^{\circ})=22\sqrt{3}=38.11 N[/tex]
Now,
x-component of resultant
[tex]R_x=F_{1x}+F_{2x}=0+22=22 N[/tex]
y-component of resultant
[tex]R_y=F_{1y}+F_{2y}=33+38.11=71.11 N[/tex]
[tex]|R|=\sqrt{R^2_x+R^2_y}[/tex]
[tex]|R|=\sqrt{(22)^2+(71.11)^2}=74.4 N[/tex]
[tex]\theta=tan^{-1}(\frac{R_y}{R_x})[/tex]
[tex]\theta=tan^{-1}(\frac{71.11}{22})=72.8^{\circ}[/tex] South
Hence, the magnitude of third force is 74.4 N and direction of third force is 72.8 degrees South.
The magnitude and direction of the third force is;
F3 = 74.44 N
θ3 = 72.81° in the south direction
We are given;
F1 = 33 N
F2 = 44 N
θ1 = 90°
θ2 = 60°
Let the third force be F3 which will serve as the resultant
Let's first find the x and y component of the forces.
F1x = F1 cos θ1
F1x = 33 × cos 90
F1x = 0 N
F1y = F1 sin θ1
F1y = 33 × sin 90
F1y = 33 N
F2x = F2 cos θ2
F2x = 44 × cos 60
F2x = 22 N
F2y = F2 sin θ2
F2y = 44 × sin 60
F2y = 38.11 N
Thus, the resultant of the x component is;
F3x = F1x + F2x
F3x = 0 + 22
F3x = 22 N
The resultant of the y component is;
F3y = F1y + F2y
F3y = 33 + 38.11
F3y = 71.11 N
Thus, magnitude of resultant of the F3 force is;
F3 = √((F3x)² + (F3y)²)
F3 = √(22² + 71.11²)
F3 = 74.44 N
The direction of the resultant of F3 is;
θ3 = tan^(-1) F3y/F3x
θ3 = tan^(-1) 71.11/22
θ3 = 72.81° in the south direction
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