Respuesta :
Answer:
Magnitude of the force is
[tex]F_3 = 2.83 [/tex]
direction of the force is given as
[tex]\theta = 45 degree[/tex] West of South
Explanation:
As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy
So here we have
[tex]\vec F_1 + \vec F_2 + \vec F_3 = 0[/tex]
here we know that first force is of magnitude 2 N towards east
[tex]\vec F_1 = 2 \hat i N[/tex]
second force is also of 2.0 N due North
[tex]\vec F_2 = 2 \hat j[/tex]
now from above equation
[tex]2\hat i + 2\hat j + \vec F_3 = 0[/tex]
[tex]\vec F_3 = -2\hat i - 2\hat j[/tex]
so magnitude of the force is given as
[tex]F_3 = \sqrt{2^2 + 2^2}[/tex]
[tex]F_3 = 2.83 [/tex]
direction of the force is given as
[tex]\theta = tan^{-1}\frac{F_y}{F_x}[/tex]
[tex]\theta = tan^{-1}\frac{-2}{-2}[/tex]
[tex]\theta = 45 degree[/tex] West of South
Answer:
F₃ is 2.83 N at 45° south of west.
Explanation:
A force, F₁, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F₂ = 2.0 N, due north.
Considering east as positive x direction and north as positive y direction.
F₁ = 2 i
F₂ = 2 j
First let us find the resultant of F₁ and F₂
[tex]F_1+F_2=2i+2j[/tex]
[tex]\texttt{Magnitude = }\sqrt{2^2+2^2}=2.83N\\\\\texttt{Direction =}tan^{-1}\frac{2}{2}=45^0[/tex]
Hence the resultant is 2.83 N at 45° north of east.
For F₃ to cancel the resultant it should be opposite to the direction of the resultant and it should have same magnitude.
So F₃ is 2.83 N at 45° south of west.
