Exercises — Pulley systems — mechanical advantage

Ask the only question that matters: how many strands of rope pull up on the load? Look at the left panel of the figure above — a single rope segment goes straight from the load up to the pulley. That's one supporting strand. $nT=W\Rightarrow 1\cdot T=W\Rightarrow \text{IMA}=n=1.$ A fixed pulley multiplies force by $1$ — it only redirects your pull (down becomes up). Handy, not powerful.

The movable pulley has the rope wrapping under it: one segment goes up to the ceiling anchor, another goes up to your hand. Both belong to the same rope, so both carry the same tension $T$, and both pull up on the movable block. Count = $2$. $2T=W\Rightarrow \text{IMA}=2.$ See the two orange strands in the right panel of the figure above.

False. A fixed pulley changes the direction of the rope but adds zero supporting strands to the movable block. MA counts only strands lifting the movable load. A pure redirector contributes nothing to $n$.

(a) Weight: $W=mg=50\times 9.8=490\text{N}$. Four strands share it equally: $4T=W\Rightarrow T=\frac{490}{4}=122.5\text{N}.$ You hold one strand, so $F_{\text{effort}}=122.5\text{N}$. (b) Why $d=nh$? The rope is inextensible (Constraint relations): its total length never changes. When the load (and its movable pulley) rise by $h$, every one of the $n$ supporting strands gets shorter by $h$. All of that removed slack — $n\times h$ in total — has to come out somewhere, and the only free end is in your hand. So your hand travels $d=nh$. Here: $d=nh=4\times 0.8=3.2\text{m}.$ Energy check: $W_{\text{in}}=122.5\times 3.2=392\text{J}$, $W_{\text{out}}=490\times 0.8=392\text{J}$. ✔ (Work–Energy Theorem)

$\text{IMA}=\frac{W}{F_{\text{effort}}}=\frac{300}{60}=5.$ Since IMA equals the number of supporting strands, $n=5$.

$W=12\times 9.8=117.6\text{N}$. Movable pulley ⇒ $n=2$. $F_{\text{effort}}=\frac{W}{2}=58.8\text{N},d=2\times 1.5=3.0\text{m}.$

One rope ⇒ one tension $T$ on both sides. Inextensible rope ⇒ same speed and same $|a|$: as $m_1$ falls by $x$, $m_2$ rises by $x$. Using the sign convention above (take down positive for the falling side $m_1$, up positive for the rising side $m_2$), free-body each mass (Newton's Second Law):

• Heavier side $m_1$ falls: $m_{1}g-T=m_{1}a$.
• Lighter side $m_2$ rises: $T-m_{2}g=m_{2}a$.

Add the equations to kill $T$: $(m_1-m_2)g=(m_1+m_2)a\Rightarrow a=\frac{(m_1-m_2)g}{m_1+m_2}=\frac{2\times 9.8}{10}=1.96{\text{m/s}}^2.$ Then from the second equation: $T=m_2(g+a)=4(9.8+1.96)=47.04\text{N}.$ Where the closed form comes from: substitute the boxed $a$ back into $T=m_2(g+a)$: $T=m_2\left(g+\frac{(m_1-m_2)g}{m_1+m_2}\right)=m_{2}g\frac{(m_1+m_2)+(m_1-m_2)}{m_1+m_2}=\frac{2m_1{m}_{2}g}{m_1+m_2}.$ That is a derived result, not a memorised one. Check: $\frac{2\cdot 6\cdot 4\cdot 9.8}{10}=47.04\text{N}$. ✔

Actual MA uses the real forces: $\text{AMA}=\frac{W}{F_{\text{effort}}}=\frac{500}{150}=3.333\dots$ $\eta =\frac{\text{AMA}}{\text{IMA}}=\frac{3.333}{4}=0.8333\dots \approx 83.3\%.$ Friction (Friction) and pulley weight ate $\approx 16.7\%$ of your work.

Using our sign convention, we take an end's rate as positive when its segment lengthens (that end feeding rope out). The rope length over the movable pulley is made of two segments, so the pulley moves at the average: $a_{\text{pulley}}=\frac{a_1+a_2}{2}=\frac{3+0}{2}=1.5{\text{m/s}}^2.$ Why average? Constant total rope length: whatever slack one end supplies, the pulley position must absorb across two segments, so it shifts at half the summed end-rate. Never guess this sign — derive it from constant length.

Speed: your hand reels in rope at $0.4\text{m/s}$. Two strands must shorten together to lift the load, so the load rises at half your hand-speed: $v_{\text{load}}=\frac{v_{\text{hand}}}{n}=\frac{0.4}{2}=0.2\text{m/s}.$ This is the same $d=nh$ trade, now per second. Power (your input): $F_{\text{effort}}=W/2=117.6/2=58.8\text{N}$, moving at $0.4\text{m/s}$: $P_{\text{in}}=F_{\text{effort}}{v}_{\text{hand}}=58.8\times 0.4=23.52\text{W}.$ Check output: $P_{\text{out}}=W{v}_{\text{load}}=117.6\times 0.2=23.52\text{W}$. ✔ Power in = power out (ideal).

Start from the definition $\eta =\frac{\text{AMA}}{\text{IMA}}$ (from Problem 3.2). Multiply both sides by $\text{IMA}$ to isolate the actual MA: $\text{AMA}=\eta \times \text{IMA}=0.9\times 3=\mathrm{2.7.}$ Now use $\text{AMA}=\frac{W}{F_{\text{effort}}}$, rearranged for the effort: $F_{\text{effort}}=\frac{W}{\text{AMA}}=\frac{270}{2.7}=100\text{N}.$ Wasted fraction $=1-\eta =1-0.9=0.1=10\%$ of input work lost to friction/pulley weight.

Step 1 — Constraint (what moves how fast). The movable pulley (with $M$) has $n=2$ strands. If $m$ descends and pulls rope at acceleration $a_m$, the load $M$ rises at half that rate: $a_M=\frac{a_m}{2}.$ Following our sign convention, let the load's upward acceleration be $a$ (up positive for $M$, which we expect to rise), so $m$ accelerates downward at $2a$ (down positive for $m$, which we expect to fall).

Step 2 — FBD of $M$ (with its movable pulley). Two strands pull up ($2T$), gravity pulls down ($Mg$). With up positive and acceleration $a$: 2T - Mg = M a. \tag{i}

Step 3 — FBD of counterweight $m$. Gravity $mg$ down, tension $T$ up, with down positive and acceleration $2a$: mg - T = m(2a). \tag{ii}

Step 4 — Solve. From (ii): $T=mg-2ma=m(g-2a)$. Substitute into (i): $2m(g-2a)-Mg=Ma$ $2mg-4ma-Mg=Ma$ $(2m-M)g=(M+4m)a$ $a=\frac{(2m-M)g}{M+4m}.$ Plug in $M=10$, $m=8$: $a=\frac{(16-10)\times 9.8}{10+32}=\frac{6\times 9.8}{42}=\frac{58.8}{42}=1.4{\text{m/s}}^2\text{ (load rises)}.$ Tension: $T=m(g-2a)=8(9.8-2.8)=8\times 7.0=56\text{N}.$ Sanity: $2T=112\text{N}>Mg=98\text{N}$, so the net upward force lifts $M$ — consistent with $a>0$. ✔

Set $a=0$ in $a=\frac{(2m-M)g}{M+4m}$: the numerator must vanish: $2m_{\star }-M=0\Rightarrow m_{\star }=\frac{M}{2}=\frac{10}{2}=5\text{kg}.$ MA interpretation: the counterweight's weight is your effort here. To balance a $10\text{kg}$ load with an $n=2$ movable pulley you need only half the load's weight, i.e. a $5\text{kg}$ counterweight. That's exactly $\text{IMA}=2$: effort $=W/2$.