Question

An object with a mass of `4` `kg` is acted on by two forces. The first is `F_1=<-4` `N, 5` `N>` and the second is `F_2 = <2` `N, -8` `N>`. What is the object's rate and direction of acceleration?

Answer 1

The object is accelerating at `0.90 "m/s"^2` in the direction `56.3^"o"` counterclockwise from negative `x`-axis.

Explanation:

The net force, `vec(F_"net")`, is given by

`vec(F_"net") = vec(F_1) + vec(F_2)`

`= < -4 "N", 5 "N" > + < 2 "N", -8 "N" >`

`= < -2 "N", -3 "N" >`

Recall Newton's `2^"nd"` law

`vec(F_"net") = m vec(a)`

where `m` is the mass of an object and `vec(a)` is its acceleration.

Substituting the values in, the resulting equation is

`< -2 "N", -3 "N" > = (4 "kg") * < a_"x" , a_"y" >`

Solving for `a_"x"` and `a_"y"`,

`a_"x" = frac{-2 "N"}{4 "kg"} = -0.5 "m/s"^2`
`a_"y" = frac{-3 "N"}{4 "kg"} = -0.75 "m/s"^2`

The object's acceleration is found to be

`vec(a) = < -0.5 "m/s"^2, -0.75 "m/s"^2 >`

Its magnitude is

`||vec(a)|| = sqrt((-0.5 "m/s"^2)^2 + (-0.75 "m/s"^2)^2)`

`= 0.90 "m/s"^2`

Its direction is

`tan^{-1}(abs(frac{-0.75 "m/s"^2}{-0.5 "m/s"^2})) = 56.3^"o"`

The angle lies in the third quadrant since both the `x` and the `y` components of the vector are negative. The direction of the acceleration is `56.3^"o"` counterclockwise from negative `x`-axis.