Answer
The magnitude for $F_3 $ is $147N$
Work Step by Step
Rearranging the whole system force equation from part (a):
$F=m_ta + μ_km_tg $
$F=m_t(a + μ_kg )$
$\frac{F}{m_t}=a + μ_kg $
Substitute the result into the eqn for $F_3$ from part (a):
$F_3=m_3(a + μ_kg )$
becomes
$F_3=m_3(\frac{F}{m_t} )$
This solution proves that $F_3$ is not dependent on the coefficient of friction.
So if it is changed the value of $F_3$ remains constant.
Therefore the magnitude for $F_3 $ is $147N$
