Question

Three forces act on a point: 3 N at 0°, 4 N at 90°, and 5 N at 217°. What is the net force?

Answer 1

The resultant force is `"1.41 N"` at `315^@`.

Explanation:

The net force `(F_"net")` is the resultant force `(F_"R")`. Each force can be resolved into an `x`-component and a `y`-component.

Find the `x`-component of each force by multiplying the force by the cosine of the angle. Add them to get the resultant `x`-component.

`Sigma(F_"x")=("3 N"*cos0^@) + ("4 N"*cos90^@) + ("5 N"*cos217^@)"="-1 "N"`

Find the `y`-component of each force by multiplying each force by the sine of the angle. Add them to get the resultant `x`-component.

`Sigma(F_y)` `=` `("3 N"*sin0^@) + ("4 N"*sin90^@) + ("5 N"*sin217^@)"="+1 "N"`

Use the Pythagorean to get the magnitude of the resultant force.

`Sigma(F_R)` `=` `sqrt((F_x)^2+(F_y)^2)`

`Sigma(F_R)` `=` `sqrt((-1 "N")^2+(1 "N")^2)`

`Sigma(F_R)` `=` `sqrt("1 N"^2 + "1 N"^2)`

`Sigma(F_R)` `=` `sqrt("2 N"^2)`

`Sigma(F_R)` `=` `"1.41 N"`

To find the direction of the resultant force, use the tangent:

`tantheta=(F_y)/(F_x)=("1 N")/(-"1 N")`

`tan^(-1)(1/(-1))=-45^@`

Subtract `45^@` from `360^@` to get `315^@`.

The resultant force is `"1.41 N"` at `315^@`.