Question

add two east-west and two north-south forces to the force diagram in order for the 3.0-kg object to have an eastward acceleration of 4.0 m/s/s.
6 n north
6 n west
8 n east
4 n south

Explanation:

Step1: Find required net force

Using Newton's second law $F_{net}=ma$, substitute $m=3.0\ \text{kg}$, $a=4.0\ \text{m/s}^2$
$F_{net} = 3.0 \times 4.0 = 12\ \text{N (East)}$

Step2: Calculate current east-west net force

Current forces: $8\ \text{N East}$, $6\ \text{N West}$
$F_{EW, current} = 8 - 6 = 2\ \text{N (East)}$

Step3: Find needed east-west force addition

Let added east-west net force be $F_{EW, add}$. $2 + F_{EW, add} = 12$
$F_{EW, add} = 12 - 2 = 10\ \text{N (East)}$
We can add, e.g., $15\ \text{N East}$ and $5\ \text{N West}$ (net $10\ \text{N East}$)

Step4: Calculate current north-south net force

Current forces: $6\ \text{N North}$, $4\ \text{N South}$
$F_{NS, current} = 6 - 4 = 2\ \text{N (North)}$

Step5: Find needed north-south force addition

To have no north-south acceleration (only eastward), net north-south force must be 0. Let added north-south net force be $F_{NS, add}$. $2 + F_{NS, add} = 0$
$F_{NS, add} = -2\ \text{N (i.e., 2 N South)}$
We can add, e.g., $3\ \text{N South}$ and $1\ \text{N North}$ (net $2\ \text{N South}$)

Answer:

1. For east-west direction: Add two forces that result in a net eastward force of 10 N (example: 15 N East and 5 N West)
2. For north-south direction: Add two forces that result in a net southward force of 2 N (example: 3 N South and 1 N North)

When added, the object will have an eastward acceleration of $4.0\ \text{m/s}^2$.