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 $D=40\ \text{in}$

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

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

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

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

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

  1. 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

  1. 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.

  1. 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).

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

  1. MA for seven strands: MA=7.

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

Inclined Plane

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

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

  1. 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}.

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

  1. 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

  1. 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.

  1. 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}.

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

Screw

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

  1. 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}.

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

  1. 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}.

  1. 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.