Notes for Simple Machines Practice Problems (PLTW)

Lever – First Class

Given: A first class lever in static equilibrium with resistance force $Fr=75\ \text{lb}$ , effort force $Fe=20\ \text{lb}$ , and the lever’s effort force located $L_e=6\ \text{ft}$ from the fulcrum.
2. Actual mechanical advantage (MA) and related values
- Formula: $MA{\text{actual}}=\frac{Fr}{F_e}=\frac{75}{20}=3.75$
- Ideal mechanical advantage (based on lever arms): $MA{\text{ideal}}=\frac{Le}{Lr}$ where $Lr$ is the resistance arm (distance from fulcrum to resistance).
- Use static equilibrium to find $L_r$:
- From moments about the fulcrum: $Fe\,Le=Fr\,Lr \Rightarrow Lr=\frac{Fe\,Le}{Fr}=\frac{20\times 6}{75}=1.6\ \text{ft}$
- Check MA with arms: $MA{\text{ideal}}=\frac{Le}{L_r}=\frac{6}{1.6}=3.75$
- Final Answer: $MA{\text{actual}}=3.75\quad (\text{which equals }MA{\text{ideal}}\text{ under ideal conditions});\ L_r=1.6\ \text{ft}.$
3. Illustration note: Sketch should label forces, distances, directions, and unknowns; all steps documented with units.

Lever – Wheelbarrow

Given: A wheelbarrow lifts a $Fr=300\ \text{lb}$ load. Distance from wheel axle to load center $Lr=3\ \text{ft}$ ; distance from wheel/axle to effort $L_e=6\ \text{ft}$ .
4. Sketch and annotate (lever system with fulcrum at wheel).
5. Ideal MA: $MA{\text{ideal}}=\frac{Le}{L_r}=\frac{6}{3}=2.0$
6. Using static equilibrium to overcome resistance: $Fe=\frac{Fr}{MA_{\text{ideal}}}=\frac{300}{2}=150\ \text{lb}$
Final Answer: $MA{\text{ideal}}=2.0;\ Fe=150\ \text{lb}.$
7. Tweezers (four-inch-long) used to squeeze; squeezing force $F_e=1\ \text{lb}$ ; if splinter experiences more than $\frac{1}{5}\ \text{lb}$ (0.2 lb) it will break.
8. Sketch and compute actual MA for tweezers
9. Using static equilibrium, calculate distance from fulcrum to splinter to avoid damage
- 7-9 are treated as a lever problem:
- The splinter resistance force limit: $Fr^{\text{max}}=0.2\ \text{lb}$ , $Fe=1\ \text{lb}$ , so
- $MA{\text{actual}}=\frac{Fr}{F_e}=\frac{0.2}{1}=0.2.$
- With lever arms $d{effort}$ (distance from fulcrum to the point where you apply the squeezing force) and $d{resistance}$ (distance from fulcrum to splinter), $MA=\frac{d{effort}}{d{resistance}}=0.2$ and $d{effort}+d{resistance}=4\ \text{in}$ (tweezer length). Solve:
- Let $d{effort}=x$, $d{resistance}=y$; $x=0.2y$ and $x+y=4$ ⇒ $1.2y=4$ ⇒ $y=3.333\text{ in}$ , $x=0.667\text{ in}$ .
- Therefore: distance from fulcrum to splinter ($d{resistance}$) ≈ $3.33\ \,\text{in}$ ; distance from fulcrum to effort point ($d{effort}$) ≈ $0.67\ \text{in}$ .
Final Answer: $MA{\text{actual}}=0.2;\ d{effort}≈0.67\in,\ d_{resistance}≈3.33\in.$

Wheel and Axle

9. Distance traveled in one revolution of a wheel with diameter $D=40\ \text{in}$

Wheel–axle system problem: valve with $Fe=40\ \text{lb}$ , resistance $Fr=250\ \text{lb}$ , axle diameter $d_{axle}=2.88\ \text{in}$ .

Sketch and annotate wheel and axle system; 11-13 solve MA and wheel diameter.

Linear distance per revolution (circumference): $C=\pi D=\pi(40\ \text{in})=40\pi\ \text{in}\approx125.66\ \text{in}.$

Required actual MA: $MA{\text{required}}=\frac{Fr}{F_e}=\frac{250}{40}=6.25.$

Required actual MA (same as above): $MA_{\text{required}}=6.25.$

Required wheel diameter to achieve this MA:

Radius of axle: $R{axle}=\dfrac{d{axle}}{2}=\dfrac{2.88}{2}=1.44\ \text{in}$ .
Using $MA=\dfrac{R{wheel}}{R{axle}}$ or $MA=\dfrac{D{wheel}}{2R{axle}}$ , we get
$D{wheel}=2\,MA\,R{axle}=2(6.25)(1.44)=18.0\ \text{in}.$

Final: $C=40\pi\text{ in}$ ; $MA{required}=6.25$; $D{wheel}=18\ \text{in}$ .

Pulley System

Construction crew lifts about $Fr=590\ \text{lb}$ from ground to 32 ft rooftop with a block and tackle that requires $Fe=45\ \text{lb}$ . What is the required actual MA?

$MA{\text{actual}}=\frac{Fr}{F_e}=\frac{590}{45}\approx13.11.$

How many supporting strands are needed?

In ideal block-and-tackle, MA equals the number of supporting strands. Since MA ≈ 13.11, you would need at least 14 strands (round up to next whole strand).

A block and tackle with seven supporting strands lifts a lathe; motor provides $F_e=150\ \text{lb}$ .

MA for seven strands: $MA=7.$

Maximum weight of lathe that can be lifted by the motor: $Fr^{\max}=MA\cdot Fe=7\times 150=1050\ \text{lb}.$

Inclined Plane

Inclined plane design: wheelchair ramp next to steps; rise $H=3.5\ \text{ft}$ ; ADA requires slope $\le 1:12$ (rise:run).

Minimum ramp base length (run): slope condition $\frac{rise}{run} \le \frac{1}{12}$ ⇒ $run \ge 12\times rise=12\times 3.5\ \text{ft}=42\ \text{ft}$ .

Using base length 42 ft, length of slope (hypotenuse): $L_{slope}=\sqrt{H^2+Run^2}=\sqrt{(3.5)^2+42^2}\text{ ft} \approx \sqrt{12.25+1764}=\sqrt{1776}\approx 42.17\ \text{ft}.$

Ideal MA for the ramp: $MA_{\text{ideal}}=\frac{Run}{Rise}=\frac{42}{3.5}=12.$

With combined weight $Fr=200\ \text{lb}$ (person + wheelchair), ideal effort required: $Fe^{\text{ideal}}=\frac{Rise}{Run}F_r=\frac{3.5}{42}\times 200=\frac{1}{12}\times200\approx 16.7\ \text{lb}.$

Wedge

Cross-section sketch: wedge cutting blade with 45° slope between the two sides; blade thickness (perpendicular distance between sides) $t=\frac{1}{5}\ \text{in}$ ; hydraulic pressure applies $F=2000\ \text{lb}$ to the wedge.

Length of the slope (along the incline) for a symmetric wedge with an angle of $45^{\circ}$ between sides:

For a wedge formed by two sides with angle $\theta$ between them, the slope length from apex to outer edge given thickness $t$ is $L_{slope}=\frac{t}{\sin(\theta/2)}.$
Here $\theta=45^{\circ}$ , so $\theta/2=22.5^{\circ}$ and $\sin(22.5^{\circ})\approx0.382683$ .
$L_{slope}=\frac{0.2}{\sin(22.5^{\circ})}=\frac{0.2}{0.382683}\approx0.5236\ \text{in}.$

Ideal MA for the wedge: $MA{\text{wedge}}=\frac{L{slope}}{t}=\frac{0.5236}{0.2}\approx 2.618.$

Screw

Screw system: 7/16 nut driver with a 1 3/4 inch diameter handle used to install a 1/4"-20 UNC bolt.

Circumference where effort is applied (handle): with $D=1\tfrac{3}{4}$ in = 1.75 in,

$C=\pi D=\pi(1.75)=1.75\pi\ \text{in}\approx5.50\ \text{in}.$

Determine the pitch of the screw (UNC 25): 25 threads per inch => pitch $p=\frac{1}{25}=0.04\ \text{in}$ per turn.

Mechanical advantage gained in the screw:

Radius of handle $R=\dfrac{D}{2}=\dfrac{1.75}{2}=0.875\ \text{in}$ ; Circumference traveled per turn $=2\pi R=2\pi(0.875)=1.75\pi\approx5.50\ \text{in}$ .
$MA_{\text{screw}}=\frac{2\pi R}{p}=\frac{5.50}{0.04}\approx\mathbf{137.4}.$

Ideals: if $F{in}=5\ \text{lb}$ is applied, the output force (overcome resistance) is $F{out}=MA{\text{screw}}\times F{in}=137.4\times 5\approx\mathbf{687.2\ \text{lb}}.$

Key takeaways and formulas (summary):
- Lever (first class): $MA{\text{actual}}=\frac{Fr}{Fe},\quad MA{\text{ideal}}=\frac{Le}{Lr},\quad Lr=\frac{FeLe}{Fr}.$
- Wheel and Axle: $MA{\text{required}}=\frac{Fr}{Fe},\quad D{wheel}=2\,MA\,R{axle},\quad R{axle}=\frac{d_{axle}}{2}.$
- Pulley: MA ≈ number of supporting strands (ideal case); MA{actual}=Fr/F_e.
- Inclined Plane: $MA_{\text{ideal}}=\frac{Run}{Rise}.$
- Wedge: $MA{\text{wedge}}=\frac{L{slope}}{t}=\frac{t}{\sin(\theta/2)}/t=\frac{1}{\sin(\theta/2)}$ for a given thickness and angle; for $\theta=45^{\circ}$ this yields $L_{slope}=\dfrac{t}{\sin 22.5^{\circ}}$ .
- Screw: $MA_{\text{screw}}=\frac{2\pi R}{p},\quad C=\pi D,\quad p=\text{pitch}.$
Notes on interpretation:
- All calculations assume ideal conditions with no friction loss, as requested.
- Where multiple distances or geometry are needed, equations were solved using the given total lengths (e.g., fixed tool length or run length) and the lever-arm relationships from static equilibrium.
- Final numeric answers are given after applying the standard equations and solving for the unknowns. When rounding, keep appropriate significant figures based on the data provided in the problem.