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\begin{document}

\frame{\frametitle{\b Newton's Laws II-1 } \bigskip

{\bf Stack of blocks (1).} \bigskip

Acceleration:
$$ F = ({\rm Total\ mass})\,a\,. $$
Net force on the first block:
$$ F_1 = m_1\,a\,.$$
Force of block 2 on block 1 (and similarly of block 3 on block 2):
$$ F21 = (m_2+m_3)\,a\,. $$
%\bigskip

{\bf Tension in a rope (2).}  Hint: the mass density of the rope is $m/L$.
\bigskip

{\bf Spider (3):} Use
$$ m\,g = 2\,T_y~~{\rm and}~~\frac{T_y}{T} = \sin(\theta)\,.$$
\bigskip
\vskip 20pt }

\frame{\frametitle{\b Newton's Laws II-2 } \bigskip

{\bf Hanging mobile (4).} 
\bigskip

Key to the problem is that the forces at each vertex have to add to
zero (Statics! Otherwise the would move). Due to symmetry is is 
sufficient to deal with one side, and we have immediately four 
equations. In Cartesian coordinates they read:
$$ T^2_x+T^1_x = 0\,,~~~T^3_x-T^2_x = 0$$
$$ T^2_y+T^1_y-m\,g = 0\,,~~~T^3_y-T^2_y-m\,g = 0$$
Where the convention is positive for the initially encountered 
components, the $x$ axis points from left to right and the 
$y$ axis up. Noting $T^1_y=0$, $T^3_y=T^3\sin(\theta)$, and 
so on the solutions are obtained.
\vskip 20pt }


\frame{\frametitle{\b Newton's Laws II-3 } \bigskip

{\bf Spring constant (5):} 
$$ F = (w_2-w_1) = \kappa\,\triangle L\,.$$

\bigskip {\bf Two Springs (6).} % \bigskip
\bigskip

Newton's 3rd law (magnitudes taken):
$$ F_1=\kappa_1\,\triangle x_1 = F_2 = \kappa_2\,\triangle x_2\,,
   ~~~\triangle x_1 + \triangle x_2 = \triangle x = d\,. $$
Solve for $y=\triangle x_1$.

\bigskip {\bf Two Spring Accelerometer (7):} % \bigskip
$$ F = m\,a_y = k\,\triangle y = k\,y\,. $$
Solve for $a_y$. Most important perhaps: This is a $k$, not a 
$\kappa$. The system will give you ``false'' if you type kappa
instead of k.
\bigskip
\vskip 20pt }

\end{document}
