F pushing right

friction pushing left

\(\displaystyle{m}_{{2}}:\)

\(\displaystyle{m}_{{1}}\) pushing right

\(\displaystyle{m}_{{3}}\) pushing left

friction pushing left

\(\displaystyle{m}_{{3}}:\)

\(\displaystyle{m}_{{2}}\) pushing right

friction pushing left

a) \(\displaystyle{F}={m}{a}\)

first we have to find the net acceleration of the system.

\(\displaystyle{F}_{{a}}-{F}_{{f}}={\left({m}_{{1}}+{m}_{{2}}+{m}_{{3}}\right)}{a}\)

\(\displaystyle{F}_{{a}}-\mu{\left({m}_{{1}}+{m}_{{2}}+{m}_{{3}}\right)}{g}={\left({m}_{{1}}+{m}_{{2}}+{m}_{{3}}\right)}{a}\)

\(\displaystyle{a}={\frac{{{F}_{{a}}-\mu{\left({m}_{{1}}+{m}_{{2}}+{m}_{{3}}\right)}}}{{{m}_{{1}}+{m}_{{2}}+{m}_{{3}}}}}\)

After you find the acceleration, multiply it by \(\displaystyle{m}_{{3}}\) and that wikk be the answer

b) \(\displaystyle{F}_{{{32}}}\) will be higher before.

In our equation, a smaller \(\displaystyle\mu\) will result in a large numerator, and a larger acceleration. Since \(\displaystyle{F}={m}{a}\), a high acceleration will also increase the force.