Question

An object with a mass of `14 kg` is acted on by two forces. The first is `F_1= < 5 N , 3 N>` and the second is `F_2 = < 2 N, 9 N>`. What is the object's rate and direction of acceleration?

Answer 1

`a = 0.993` `"m/s"^2`

`theta = 59.7^"o"`

Explanation:

We're asked to find the magnitude and direction of the acceleration of an object, given its mass and two forces that act on it.

To do this, we'll split the problem into `x`- and `y`-components, so we have

`F_(1x) = 5` `"N"`

`F_(1y) = 3` `"N"`

`F_(2x) = 2` `"N"`

`F_(2y) = 9` `"N"`

We can add respective components to find the net force components:

`sumF_x = F_(1x) + F_(2x) = 5` `"N"` `+ 2` `"N"` `= color(red)(7color(white)(l)"N"`

`sumF_y = F_(1y) + F_(2y) = 3` `"N"` `+ 9` `"N"` `= color(green)(12color(white)(l)"N"`

Now, we use Newton's second law to find the components of the object's acceleration:

`a_x = (sumF_x)/m = color(red)(7color(white)(l)"N")/(14color(white)(l)"kg") = color(purple)(0.5color(white)(l)"m/s"^2`

`a_y = (sumF_y)/m = color(green)(12color(white)(l)"N")/(14color(white)(l)"kg") = color(orange)(6/7color(white)(l)"m/s"^2`

The magnitude of the acceleration is thus

`a = sqrt((a_x)^2 + (a_y)^2) = sqrt((color(purple)(0.5color(white)(l)"m/s"^2))^2 + (color(orange)(6/7color(white)(l)"m/s"^2))^2`

`= color(blue)(0.993` `color(blue)("m/s"^2`

and the direction is

`theta = arctan((a_y)/(a_x)) = arctan((color(orange)(6/7color(white)(l)"m/s"^2))/(color(purple)(0.5color(white)(l)"m/s"^2))) = color(blue)(59.7^"o"`