Question

An object with a mass of `4 kg` is acted on by two forces. The first is `F_1= < 6 N , -2 N>` and the second is `F_2 = < -3 N, 3 N>`. What is the object's rate and direction of acceleration?

Answer 1

The acceleration, `vec(a) = 1/4[(3), (1)]m/s^2`
and the magnitude is:
`|vec(a)| = 1/4sqrt[(3)^2+(1)^2]N=1/4sqrt(10)m/s^2`
You can write the acceleration, `vec(a)= sqrt10/4[(3/sqrt(10)), (1/sqrt(10))]m/s^2`
Where`[(3/sqrt(10)), (1/sqrt(10))]=` is the unit vector of `vec(a)`

Explanation:

Given: Forces on an object of mass, `m=4kg`
`vec(F_1)=[(6), (-2)]N` and `vec(F_2)=[(-3), (3)]N`

Required: Object acceleration, `vec(a)`

Solution Strategy:
A) Draw a "Free Body Diagram" and add the two force vector
B) Use Newton's law `sum_l(F_xhati + F_yhatj)=m(a_xhati+a_yjhat)`
That is:
1) `" "sum_l F_x = ma_x" "` `and" "` 2) `" "sum_l F_y = ma_y`

A) See Image Below

The FBD yield:
`F_(x)=F_(1_x)+F_(2_x)=6-3= 3N`
`F_(y)=F_(1_y)+F_(2_y)=-2+3= 1N`
Thus the net force, `vec(F_"net") = [(3), (1)]N`

B) Newton Law:
From equation 1) we have:
1) `F_(x)=ma_x; a_x=3/4 m/s^2`
`F_(y)=ma_y; a_y=1/4 m/s^2`

Thus the net acceleration, `vec(a) = 1/4[(3), (1)]m/s^2`
and the magnitude is:
`|vec(a)| = 1/4sqrt[(3)^2+(1)^2]N = 1/4sqrt(10)m/s^2`