Exercises — Tension in inextensible strings

WHAT the picture shows: one block, one string, nothing moving. WHY static: "at rest" means acceleration $a=0$. FBD of the block: two vertical forces — weight $mg$ pulling down, tension $T$ pulling up (a string always pulls the body toward the string, i.e. upward here). $T-mg=ma=0\Rightarrow T=mg$ $T=3\times 10=\boxed{30\text{N}}$ The static shortcut "$T=mg$" is legal only because $a=0$. Remember that — the very next level breaks it.

Why the accelerations must match — step by step. Call the two string lengths hanging down from the pulley $x_1$ (to block 1) and $x_2$ (to block 2). The string can't stretch, so the total length is fixed: $x_1+x_2=\text{constant}.$ Nothing on the right-hand side changes with time, so its rate of change is zero. Differentiate once with respect to time (rate of change of position = velocity): ${\dot{x}}_1+{\dot{x}}_2=0\Rightarrow {\dot{x}}_1=-{\dot{x}}_2(\text{one end pays out exactly what the other reels in}).$ Differentiate again (rate of change of velocity = acceleration): ${\ddot{x}}_1=-{\ddot{x}}_2.$ Now define the symbol $a$ to be the common magnitude of these two accelerations — the single positive number that both blocks share. In symbols, $a\equiv |{\ddot{x}}_1|=|{\ddot{x}}_2|$, i.e. $|a_1|=|a_2|=a.$ The minus sign above is the geometry: one goes down while the other goes up, but the magnitudes are locked equal, and $a$ is the name we give that shared magnitude.

• (a) Wrong — weights depend on the masses, which differ.
• (b) Correct — $|a_1|=|a_2|=a$.
• (c) Wrong — over a pulley one goes down while the other goes up; directions are opposite. Answer: (b).

This is the canonical Atwood Machine. In the figure the red block is $m_1$ (the heavier one on the left), and you can see its downward motion arrow while the white $m_2$ has an upward arrow — the picture is telling you which way to point "positive" for each body. Rather than blindly quoting the formula, let's rebuild it from the two FBDs so the shape of the answer makes sense. FBD of $m_1$ (heavier red block, falls; positive = down): weight $m_{1}g$ down, tension $T$ up. $m_{1}g-T=m_{1}a$ FBD of $m_2$ (lighter white block, rises; positive = up): tension $T$ up, weight $m_{2}g$ down. $T-m_{2}g=m_{2}a$ Why add them? Adding cancels the unknown $T$ and leaves only $a$: $m_{1}g-m_{2}g=(m_1+m_2)a\Rightarrow a=\frac{(m_1-m_2)g}{m_1+m_2}.$ The numerator is the net driving weight $(m_1-m_2)g$; the denominator is the total mass being moved — that's why the formula looks like (unbalanced force)/(total mass), which is just $F=ma$. $a=\frac{(5-3)\times 10}{5+3}=\frac{20}{8}=\boxed{2.5{\text{m/s}}^2}$ Back-substitute into $T-m_{2}g=m_{2}a$ to get tension: $T=m_2(g+a)=3(10+2.5)=\boxed{37.5\text{N}}$ (Equivalently $T=\frac{2m_1{m}_2}{m_1+m_2}g=\frac{2\cdot 5\cdot 3}{8}\times 10=37.5\text{N}$.) Sanity check: $T=37.5\text{N}$ sits between $m_{2}g=30\text{N}$ and $m_{1}g=50\text{N}$ — exactly where a real pulling force should lie. ✓

Again, build it from FBDs so we see why only $m_2$'s weight drives the system. FBD of $m_2$ (hangs, falls; positive = down): weight $m_{2}g$ down, tension $T$ up. $m_{2}g-T=m_{2}a$ FBD of $m_1$ (slides horizontally; positive = toward pulley): the only horizontal force is $T$ (weight and normal force cancel vertically, see Normal Force). $T=m_{1}a$ Add the two to cancel $T$: $m_{2}g=(m_1+m_2)a\Rightarrow a=\frac{m_{2}g}{m_1+m_2}=\frac{6\times 10}{10}=\boxed{6{\text{m/s}}^2}$ Notice the numerator is $m_{2}g$ alone — that's why the hanging mass, not the table mass, sets the driving force: only $m_2$ has gravity pulling along the string. Then from $T=m_{1}a$: $T=m_{1}a=4\times 6=\boxed{24\text{N}}$ Note $T=24\text{N}<m_{2}g=60\text{N}$: because $m_2$ is falling, the string can't be holding its full weight.

In the figure, the red arrow on the right is the applied force $F$, and notice it touches only the front block — that single visual fact is the key to the whole solution: back blocks feel string tension, never $F$ directly. The labels $T_2$ (between the two rear blocks) and $T_1$ (between the front pair) mark the two strings we must find. Step 1 — Whole system for $a$. Internal tensions cancel across the whole chain, so only $F$ acts on the total mass: $F=(m_1+m_2+m_3)a\Rightarrow a=\frac{12}{1+2+3}=\frac{12}{6}=\boxed{2{\text{m/s}}^2}$ Step 2 — Isolate the back block $m_3$ for $T_2$. The only horizontal force on $m_3$ is the string tension $T_2$ pulling it forward: $T_2=m_{3}a=3\times 2=\boxed{6\text{N}}$ Step 3 — Isolate $\{m_2,m_3\}$ together for $T_1$. The only external forward pull on this pair is $T_1$: $T_1=(m_2+m_3)a=5\times 2=\boxed{10\text{N}}$ Check: front block FBD gives $F-T_1=m_{1}a\Rightarrow 12-10=1\times 2=2$ ✓. Tensions get smaller toward the back because each string only has to accelerate the mass behind it.

FBD: weight $mg$ down, tension $T$ up. Pulled up (positive = up): $T-mg=ma\Rightarrow T=m(g+a)=2(10+3)=\boxed{26\text{N}}$ Lowered, accelerating down (positive = down): $mg-T=ma\Rightarrow T=m(g-a)=2(10-3)=\boxed{14\text{N}}$ Same bucket, same string, two different tensions — because tension responds to acceleration, not just weight. Note $26>mg>14$: speeding up needs more than weight, easing down needs less.

In the figure the red block is $m_1$ on the incline; its motion arrow points up the slope, and the hanging white $m_2$ has a downward arrow — that is the direction we will now justify with numbers. The angle $\theta =30^{\circ }$ marked at the base of the triangle is what we feed into $\sin \theta$. The new ingredient: on the incline, gravity splits into a part along the slope, $m_{1}g\sin \theta$, and a part pressing into the slope, $m_{1}g\cos \theta$. Only the along-slope part competes in the tug-of-war; the into-slope part is cancelled by the Normal Force. We need $\sin \theta$ because it is exactly the fraction of gravity that lies along the direction of motion. $m_{1}g\sin \theta =4\times 10\times \frac{1}{2}=20\text{N},m_{2}g=30\text{N}$ Since $m_{2}g=30>20$, the hanging block wins and falls; $m_1$ slides up the incline (matching the arrows in the figure). FBDs (positive = direction of motion for each):

• Hanging $m_2$ (falls): $m_{2}g-T=m_{2}a$
• Incline $m_1$ (up the slope): $T-m_{1}g\sin \theta =m_{1}a$ Add to cancel $T$: $m_{2}g-m_{1}g\sin \theta =(m_1+m_2)a$ $a=\frac{(m_2-m_1\sin \theta )g}{m_1+m_2}=\frac{(3-4\cdot 0.5)\times 10}{7}=\frac{10}{7}\approx \boxed{1.43{\text{m/s}}^2}$ Back-substitute for $T$: $T=m_2(g-a)=3\left(10-\frac{10}{7}\right)=3\cdot \frac{60}{7}=\frac{180}{7}\approx \boxed{25.7\text{N}}$ Check ranges: $20\text{N}<T\approx 25.7\text{N}<30\text{N}$ — the tension sits between the two competing pulls, as it must. ✓

Reuse the derived formula, keeping the sign convention "positive = $m_2$ falls / $m_1$ up-slope": $a=\frac{(m_2-m_1\sin \theta )g}{m_1+m_2}=\frac{(1.5-4\cdot 0.5)\times 10}{4+1.5}=\frac{(1.5-2)\times 10}{5.5}=\frac{-5}{5.5}\approx \boxed{-0.91{\text{m/s}}^2}$ The negative sign is the physics talking: our assumed direction was wrong. In reality $m_1$ slides down the incline and $m_2$ rises, with magnitude $|a|\approx 0.91{\text{m/s}}^2$. Tension. Tension is a pulling force along the string, and its strength (the number of newtons) is a single value that does not depend on which way we chose to call positive — flipping our sign convention would flip the sign of $a$ but leave the actual pull unchanged. So we may substitute the signed value $a\approx -0.91{\text{m/s}}^2$ straight into $m_2$'s equation and read off the strength: $T=m_2(g-a)=1.5\left(10-(-0.91)\right)=1.5\times 10.91\approx \boxed{16.4\text{N}}$ This is why we keep the algebraic sign instead of guessing the direction: the equations tell us the truth automatically.

Equal masses ⇒ $a=\frac{(m_1-m_2)g}{m_1+m_2}=0$, so the system is static: $T=m_{1}g=5\times 10=\boxed{50\text{N}}\text{on both sides}.$ Why equal: a massless string element obeys $T_2-T_1=(dm)a=0$, so tension can't change from point to point — including across a frictionless, massless pulley, which has no rotational inertia to create a difference. See Frictionless Pulleys vs Pulleys with Inertia.

Start from $a=\frac{(m_1-m_2)g}{m_1+m_2}$ and $T=\frac{2m_1{m}_2}{m_1+m_2}g$.

(a) $m_1=m_2=m$: substitute equal masses: $a=\frac{(m-m)g}{m+m}=\frac{0}{2m}g=\boxed{0},T=\frac{2m\cdot m}{m+m}g=\frac{2m^2}{2m}g=\boxed{mg}.$ Balanced — each side merely holds its own weight, no motion. ✓

(b) $m_2=0$: substitute zero for the second mass: $a=\frac{(m_1-0)g}{m_1+0}=\frac{m_{1}g}{m_1}=\boxed{g},T=\frac{2m_1\cdot 0}{m_1+0}g=\boxed{0}.$ Nothing on the other end ⇒ the remaining mass is in free fall, and with no mass to pull the string goes slack (zero pull). ✓

(c) $m_1\to \infty$ (with $m_2$ fixed): divide the top and bottom of each formula by $m_1$, then let $m_2/m_1\to 0$: $a=\frac{(1-m_2/m_1)g}{1+m_2/m_1}\to \frac{(1-0)g}{1+0}=\boxed{g},T=\frac{2m_2}{1+m_2/m_1}g\to \frac{2m_2}{1+0}g=\boxed{2m_{2}g}.$ A limitlessly heavy $m_1$ falls at $g$, and it yanks the light mass upward so hard the string tension approaches $2m_{2}g$ — that is the light mass's weight $m_{2}g$ plus the extra force $m_{2}g$ needed to accelerate it upward at $g$. ✓

Static single hanger, $T=?$ ::: $T=mg$ (only because $a=0$). Which quantity is shared by masses on one inextensible string? ::: acceleration magnitude $a$. Chain of blocks pulled by $F$: which string has the largest tension? ::: the one nearest the pull (it accelerates the most mass behind it). Bucket accelerated up at $a$, tension? ::: $T=m(g+a)$, more than $mg$. Bucket lowered at $a$, tension? ::: $T=m(g-a)$, less than $mg$. On an incline, what fraction of gravity drives motion? ::: the along-slope part $mg\sin \theta$. A negative computed $a$ means…? ::: the true motion is opposite to your assumed positive direction. Atwood limit $m_1\to \infty$: $a$ and $T$? ::: $a\to g$, $T\to 2m_{2}g$.