AP Physics 1 : Introduction: Forces in One Dimension Study Notes

An important feature of Newton’s second law is that it can be applied to an entire system, or to any component of the system.

For the Atwood’s machine pictured, we found that:
Entire system:
$
\begin{aligned}
\overrightarrow{\boldsymbol{F}}_{\text {net }} & =m \overrightarrow{\boldsymbol{a}} \\
+50 \mathrm{~N} & =(5 \mathrm{~kg}+10 \mathrm{~kg})\left(3 . \overline{3} \frac{\mathrm{m}}{\mathrm{s}^2}\right)
\end{aligned}
$
We can apply Newton’s second law to each block separately:
$
\begin{aligned}
& \overrightarrow{\boldsymbol{F}}_{1, n e t}=m_1 \overrightarrow{\boldsymbol{a}} \\
& \overrightarrow{\boldsymbol{F}}_{2, n e t}=m_2 \overrightarrow{\boldsymbol{a}}
\end{aligned}
$
Because the blocks are connected via the same rope, the acceleration is the same for both blocks.
This means that we can apply Newton’s second law to either of the blocks to determine the tension in the rope:

$\begin{gathered}\overrightarrow{\boldsymbol{F}}_{1, \text { net }}=\left(\overrightarrow{\boldsymbol{F}}_T-\overrightarrow{\boldsymbol{F}}_{g, 1}\right) \quad \overrightarrow{\boldsymbol{F}}_{1, \text { net }}=m_1 \overrightarrow{\boldsymbol{a}} \\ \left(\overrightarrow{\boldsymbol{F}}_T-m_1 \overrightarrow{\boldsymbol{g}}\right)=m_1 \overrightarrow{\boldsymbol{a}} \\ {\left[\overrightarrow{\boldsymbol{F}}_T-(5)(10)\right]=(5)(3 . \overline{3})} \\ \overrightarrow{\boldsymbol{F}}_T-(-50)=16 . \overline{6} \\ \overrightarrow{\boldsymbol{F}}_T=66 . \overline{6} \mathrm{~N}\end{gathered}$

Block \#2: (same calculation; yields same result)
We can do the same calculation for Block \#2, with the same result for $\overrightarrow{\boldsymbol{F}}_T$. (Remember that we chose the positive direction to be the direction that the system moves. This means the positive direction is up for block \#1, but down for block \#2.)

$\begin{gathered}\overrightarrow{\boldsymbol{F}}_{2, n e t}=\left(\overrightarrow{\boldsymbol{F}}_{g, 2}-\overrightarrow{\boldsymbol{F}}_T\right) \quad \overrightarrow{\boldsymbol{F}}_{2, \text { et }}=m_2 \overrightarrow{\boldsymbol{a}} \\ \left(m_2 \overrightarrow{\boldsymbol{g}}-\overrightarrow{\boldsymbol{F}}_T\right)=m_2 \overrightarrow{\boldsymbol{a}} \\ {\left[(10)(10)-\overrightarrow{\boldsymbol{F}}_T\right]=(10)(3 . \overline{3})} \\ 100-\overrightarrow{\boldsymbol{F}}_T=33 . \overline{3} \\ \overrightarrow{\boldsymbol{F}}_T=66 . \overline{6} \mathrm{~N} \\ \text { Q.E.D. }\end{gathered}$

A variation of Atwood’s machine is to have one of the masses on a horizontal table (possibly on a cart to reduce friction). This means that the net force is only the action of gravity on the hanging mass.

Consider the problem illustrated to the right. To simplify the problem, we will assume that the pulley has negligible mass, and that both the pulley and the cart are frictionless. The $10 \mathrm{~kg}$ for the mass on the left includes the mass of the cart.

The forces on the two masses are:

Gravity and the normal force cancel for the cart. The tensions cancel because they are equal (it’s the same rope) and are in opposite directions. This means that the only uncancelled force is the force of gravity on the $4 \mathrm{~kg}$ mass. This uncancelled force is the net force, which is $\overrightarrow{\boldsymbol{F}}_{n e t}=m g=(4)(10)=40 \mathrm{~N}$.
The total mass is $10+4=14 \mathrm{~kg}$.
Now that we have the net force and the total mass, we can find the acceleration using Newton’s Second Law:
$
\begin{aligned}
\overrightarrow{\boldsymbol{F}}_{n e t} & =m \overrightarrow{\boldsymbol{a}} \\
40 & =14 \overrightarrow{\boldsymbol{a}} \\
\overrightarrow{\boldsymbol{a}} & =\frac{40}{14}=2.86 \frac{\mathrm{m}}{\mathrm{s}^2}
\end{aligned}
$

If we also need to find the tension, we can apply Newton’s second law to the cart:
$
\begin{aligned}
& F_{\text {net, }, \text { art }}=F_T \\
& m_{\text {cart }} a=F_T \\
& (10)(2.86)=28.6 \mathrm{~N}
\end{aligned}
$
or we can apply Newton’s second law to the hanging mass:
$
\begin{aligned}
& F_{\text {net, hang }}=F_g-F_T \\
& m_{\text {hang }} a=F_g-F_T \\
& (4)(2.86)=(4)(10)-F_T \\
& 11.4=40-F_T \\
& F_T=40-11.4=28.6 \mathrm{~N}
\end{aligned}
$

Alternative Approach
In most physics textbooks, the solution to Atwood’s machine problems is presented as a system of equations. The strategy is:
– Draw a free-body diagram for each block.
– Apply Newton’s $2^{\text {nd }}$ Law to each block separately, giving $F_{\text {net }}=m_1 a$ for block 1 and $F_{\text {net }}=F_g-F_T=m_2 a$, which becomes $F_{\text {net }}=m_2 g-F_T=m_2 a$ for block 2 .
– Set the two $F_{n e t}$ equations equal to each other, eliminate one of $F_T$ or $a$, and solve for the other.

This is really just a different presentation of the same approach, but most students find it less intuitive.