Question

An object with a mass of `9 kg` is acted on by two forces. The first is `F_1= < -7 N , 1 N>` and the second is `F_2 = < 4 N, -5 N>`. What is the objects rate and direction of acceleration?

Answer 1

`a = 5/9` `"m/s"^2`

`theta = 233^"o"`

Explanation:

We're asked to find the acceleration (magnitude and direction) of an object with a known mass and two forces that act on it.

We can use the famous equation

`sumvecF = mveca`

to help solve this problem. In component form, this becomes

`sumF_x = ma_x`

`sumF_y = ma_y`

We therefore need to find the components of the net force acting on the object.

The net force components are the vector sums of the components of the individual forces:

`sumF_x = -7` `"N"` `+ 4` `"N"` `= -3` `"N"`

`sumF_y = 1` `"N"` `- 5` `"N"` `= -4` `"N"`

The components of the acceleration are

`a_x = (sumF_x)/m = (-3color(white)(l)"N")/(9color(white)(l)"kg") = -1/3` `"m/s"^2`

`a_y = (sumF_y)/m = (-4color(white)(l)"N")/(9color(white)(l)"kg") = -4/9` `"m/s"^2`

The magnitude of the acceleration is thus

`a = sqrt((a_x)^2 + (a_y)^2) = sqrt((-1/3color(white)(l)"m/s"^2)^2 + (-4/9color(white)(l)"m/s"^2)^2)`

`= color(red)(5/9` `color(red)("m/s"^2`

The direction of the acceleration is

`theta = arctan((a_y)/(a_x)) = arctan((-4/9cancel("m/s"^2))/(-1/3cancel("m/s"^2))) = 53.1^"o" + 180^"o"`

`= color(blue)(233^"o"`

(The `180^"o"` was added because the acceleration is directed toward the third quadrant relative to the object.)