Respuesta :
Answer:
The x-component of the net external force Fx = 88.0 N î
The y-component of the net external force Fy = - 44 N ĵ
Explanation:
Net force acting on an object is the sum of all forces acting upon it.
F = F1 + F2 + F3
Forces in terms of x,y-components:
F₁ = 40 N î (where î is the unit vector in x-direction)
F₂ = -80 N ĵ (where ĵ is the unit vector in y-direction)
F₃ = 60 N
= F₃cosθ î + F₃sinθ ĵ
= 60*cos(36.9°) î + 60*sin(36.9°) ĵ
= 48î + 36ĵ
Net External Force = F1 + F2 + F3
= 40 î - 80 ĵ + 48 î + 36 ĵ
= 40 î + 48 î - 80 ĵ + 36 ĵ
= 88 î - 44 ĵ
Here, only like components(same direction) can be added into each other.
Force is explained as the multiplication of mass and acceleration. Its unit is N.
The x-component of the net external force will be ( Fx )= 88.0 N î.
The y-component of the net external force will be (Fy ) = - 44 N ĵ
What is force?
Force is explained as the multiplication of mass and acceleration. Its unit is N. It is the external agent which helps to change the shape and size of an object.
Force is used to pushing and pulling the object and sometimes changes the direction of motion.
The net force acting on an object is equal to the sum of all forces acting upon it.
F = F1 + F2 + F3
F1 = 40.0 N in the +x-direction= [tex]40 \hat{i}[/tex]
F2 = - 80.0 N in the -y-direction= [tex]-81 \hat{j}[/tex]
F3 = 60.0 N
The force F₃ is at an angle of 36.9⁰counterclockwise from the +x-direction
So that
F₃ according to the cartesian coordinate will be
[tex]F_3=F_3cos\theta\:\hat{i} + F_3sin\theta \: \hat{j}[/tex]
[tex]F_3=60cos36.39^0\:\hat{i} + 60sin36.9^0 \: \hat{j}[/tex]
[tex]F_3=48\hat{i} + 36\hat{j}[/tex]
[tex]\vec{F}= \vec{F_1}+ \vec{F_2}+\vec{F_3}[/tex]
[tex]\vec{F}= 40\vec{i} + 48 \vec{i} - 80 \vec{j} + 36 \vec{j}\\\\\vec{F}= 88 \vec{i} - 44 \vec{j}[/tex]
Hence the x-component of the net external force will be ( Fx )= 88.0 N î and the y-component of the net external force will be (Fy ) = - 44 N ĵ
To learn more about the force refer to the link;
https://brainly.com/question/26115859
