This is very difficult to explain without free-bodydiagrams (FBD). I'll try to draw something:
The stuff on the right is supposed to be a spring.
The FBD for the top mass:
The bottom mass:
\(\displaystyle{f}=\mu{n}=\mu{m}{g}\)
F is the force due to the spring (F=kx). Applying Newton's 2nd law to the first FBD:
\(\displaystyle\mu{m}{g}={m}{a}\)
And to the second FBD:
\(\displaystyle{k}{x}-\mu{\left({2}{m}\right)}{g}={m}{a}\)
The blocks don't slide when they have the same acceleration:
\(\displaystyle{a}=\mu{g}={\frac{{{k}{x}}}{{{m}}}}-{2}\mu{g}\)
\(\displaystyle\mu={\frac{{{k}{x}}}{{{3}{m}{g}}}}\)
m is the mass of a single block, calculated in the first partof the problem. The blocks will slide when x = 0.05; plugthis value for x into the equation and solve.
The stuff on the right is supposed to be a spring.
The FBD for the top mass:
The bottom mass:
\(\displaystyle{f}=\mu{n}=\mu{m}{g}\)
F is the force due to the spring (F=kx). Applying Newton's 2nd law to the first FBD:
\(\displaystyle\mu{m}{g}={m}{a}\)
And to the second FBD:
\(\displaystyle{k}{x}-\mu{\left({2}{m}\right)}{g}={m}{a}\)
The blocks don't slide when they have the same acceleration:
\(\displaystyle{a}=\mu{g}={\frac{{{k}{x}}}{{{m}}}}-{2}\mu{g}\)
\(\displaystyle\mu={\frac{{{k}{x}}}{{{3}{m}{g}}}}\)
m is the mass of a single block, calculated in the first partof the problem. The blocks will slide when x = 0.05; plugthis value for x into the equation and solve.
