Question

An object with a mass of `4 kg` is acted on by two forces. The first is `F_1= < -5 N , 2 N>` and the second is `F_2 = < 7 N, 6 N>`. What is the object's rate and direction of acceleration?

Answer 1

`a = 2.06` `"m/s"^2`

`theta = 76.0^"o"`

Explanation:

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

To do this, let's first split up the forces into their components:

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

`F_(1y) = 2` `"N"`

`F_(2x) = 7` `"N"`

`F_(2y) = 6` `"N"`

We'll now find the components of the net force that acts on the body, by adding components:

`sumF_x = -5` `"N"` `+ 7` `"N"` `= 2` `"N"`

`sumF_y = 2` `"N"` `+ 6` `"N"` `= 8` `"N"`

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

`sumF_x = ma_x`

`a_x = (sumF_x)/m = (2color(white)(l)"N")/(4color(white)(l)"kg") = 0.5` `"m/s"^2`

`a_y = (sumF_y)/m = (8color(white)(l)"N")/(4color(white)(l)"kg") = 2` `"m/s"^2`

The magnitude of the acceleration is

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

`= color(red)(2.06` `color(red)("m/s"^2`

And the direction is

`theta = arctan((a_y)/(a_x)) = arctan((2cancel("m/s"^2))/(0.5cancel("m/s"^2))) = color(blue)(76.0^"o"`

Always be sure to check your direction calculation, as it could be `180^"o"` off! (by noting what direction the acceleration is relative to the object).