Question

Question #4462e

Answer 1

`F_"net" = 17.0` `"lb"`

Explanation:

We're asked to find the net force acting on the object, given two constituent forces.

We're not given which force is which, but this won't matter for finding the magnitude of the resultant force.

Let's find the components of each force vector:

• `F_(1x) = (12color(white)(l)"lb") cos (72.5^"o") = 3.61color(white)(l)"lb"`

• `F_(1y) = (12color(white)(l)"lb") sin (72.5^"o") = 11.4color(white)(l)"lb"`

• `F_(2x) = (5.66color(white)(l)"lb")cos(72.5^"o" + 30.0^"o") = -1.70color(white)(l)"lb"`

• `F_(2y) = (5.66color(white)(l)"lb")sin(72.5^"o" + 30.0^"o") = 5.40color(white)(l)"lb"`

The components of the net force are given by

• `F_x = F_(1x) + F_(2x) = 3.61color(white)(l)"lb" - 1.70color(white)(l)"lb" = color(red)(ul(1.91color(white)(l)"lb"`

• `F_y = F_(1y) + F_(2y) = 11.4color(white)(l)"lb" + 5.40color(white)(l)"lb" = color(green)(ul(16.8color(white)(l)"lb"`

The magnitude of the net force is

`color(blue)(F_"net") = sqrt((F_x)^2 + (F_y)^2) = sqrt((color(red)(1.91color(white)(l)"lb"))^2 + (color(green)(16.8color(white)(l)"lb"))^2) = color(blue)(ulbar(|stackrel(" ")(" "17.0color(white)(l)"lb"" ")|)`

You could find the direction the net force makes with the horizontal using the relation

`tantheta = (F_y)/(F_x)`

The direction can be one of two angles because we're not given which force was which direction-wise.