Question

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. What is the magnitude and direction of a third force, F3, which must be exerted on the object so that the resultant force is zero?

Answer 1

Answer:

Magnitude of the force is

$F_3 = 2.83$

direction of the force is given as

$\theta = 45 degree$ 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

$\vec F_1 + \vec F_2 + \vec F_3 = 0$

here we know that first force is of magnitude 2 N towards east

$\vec F_1 = 2 \hat i N$

second force is also of 2.0 N due North

$\vec F_2 = 2 \hat j$

now from above equation

$2\hat i + 2\hat j + \vec F_3 = 0$

$\vec F_3 = -2\hat i - 2\hat j$

so magnitude of the force is given as

$F_3 = \sqrt{2^2 + 2^2}$

$F_3 = 2.83$

direction of the force is given as

$\theta = tan^{-1}\frac{F_y}{F_x}$

$\theta = tan^{-1}\frac{-2}{-2}$

$\theta = 45 degree$ West of South

Answer 2

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

$F_1+F_2=2i+2j$

$\texttt{Magnitude = }\sqrt{2^2+2^2}=2.83N\\\\\texttt{Direction =}tan^{-1}\frac{2}{2}=45^0$

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.