Question

add two east-west and two north-south forces to the force diagram in order for the 2.0-kg object to have a westward acceleration of 4.0 m/s/s.
2 n north
6 n west
4 n east
4 n south

Explanation:

Step1: Calculate net required force

Use $F_{net}=ma$.
$F_{net}=2.0\ \text{kg} \times 4.0\ \text{m/s}^2 = 8.0\ \text{N west}$

Step2: Find current east-west net force

Subtract east from west force.
$F_{current,ew}=6\ \text{N west} - 4\ \text{N east} = 2\ \text{N west}$

Step3: Find needed east-west force addition

Subtract current net from required net.
$F_{add,ew}=8\ \text{N west} - 2\ \text{N west} = 6\ \text{N west}$
We can split this into two east-west forces: e.g., add $7\ \text{N west}$ and $1\ \text{N east}$ (net $6\ \text{N west}$; any pair with net $6\ \text{N west}$ works).

Step4: Balance north-south forces

Current north-south net force: $4\ \text{N south} - 2\ \text{N north} = 2\ \text{N south}$. To have no north-south acceleration (since acceleration is only westward), add forces with net $2\ \text{N north}$.
Example: add $3\ \text{N north}$ and $1\ \text{N south}$ (net $2\ \text{N north}$; any pair with net $2\ \text{N north}$ works).

Answer:

1. East-west forces example: Add $\boldsymbol{7\ \text{N West}}$ and $\boldsymbol{1\ \text{N East}}$ (net westward addition of $6\ \text{N}$ to reach required net force).
2. North-south forces example: Add $\boldsymbol{3\ \text{N North}}$ and $\boldsymbol{1\ \text{N South}}$ (balances existing north-south net force to ensure no north-south acceleration).

(Note: Any two east-west forces with a net 6 N west, and any two north-south forces with a net 2 N north, are valid solutions.)