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

F₃ is 2.83 N at 45° south of west.

Step-by-step 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 = }√(2^2+2^2)=2.83N\\\\\texttt{Direction =}tan^(-1)(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.

Answer 2

Magnitude of the force is

`F_3 = 2.83`

direction of the force is given as

`\theta = 45 degree` West of South

Step-by-step 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 = √(2^2 + 2^2)`

`F_3 = 2.83`

direction of the force is given as

`\theta = tan^(-1)(F_y)/(F_x)`

`\theta = tan^(-1)(-2)/(-2)`

`\theta = 45 degree` West of South