6.1 Balance Scales (Teraziler), Symbolic Weight Substitution & Inequality Scales

Balance Scales (Teraziler), Symbolic Weight Substitution & Inequality Scales

Quick Method: Treat every balanced scale (dengede olan terazi) as an algebraic equation where the left pan (sol kefe) equals the right pan (sağ kefe), and every tilted scale (dengede olmayan terazi) as a strict inequality where the lower pan is heavier (aşağıdaki kefe daha ağır). Before solving for individual shapes (●, ■, ▲), check whether canceling matching shapes on opposite pans or adding two scales pan-by-pan directly builds the target expression.

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1. Equilibrium Pan Rules (Dengede Olan Teraziler)

When a scale beam is horizontal and the central pointer stands upright, the two pans are in static equilibrium (denge durumu). Every identical symbol represents a strictly positive weight ($W > 0$).

Rule 1: Direct Cancelation on Opposite Pans (Karşılıklı Sadeleştirme)

If the exact same shape appears on both the left pan and the right pan of a single balanced scale, remove it from both sides immediately:

$$2X + Y = X + 3Z \implies X + Y = 3Z$$

Similarly, if every pan entry is a multiple of an integer $k$, divide both pans by $k$. For example, if $4C = 2S + 2T$, dividing by $2$ immediately gives $2C = S + T$.

Rule 2: Pan-by-Pan Addition and Subtraction (Taraf Tarafa Toplama ve Çıkarma)

When two balanced scales are given:

• Scale 1: $L_1 = R_1$
• Scale 2: $L_2 = R_2$

you can combine both left pans onto one side and both right pans onto the other side, or add $L_1 + R_2 = R_1 + L_2$, and the combined scale remains balanced:

$$L_1 + L_2 = R_1 + R_2$$

Rule 3: LCM Parametric Weight Assignment (EKOK ile Değer Verme)

Most classic TR-YÖS scale items provide two balanced scales containing three unknown shapes ($N_{\text{vars}} = N_{\text{eqs}} + 1$). Because all shape weights scale proportionally, you do not need to carry symbolic fractions. Instead, express two variables in terms of a single reference weight and assign the Least Common Multiple (EKOK) of the denominators.

Suppose Scale 1 gives $2X = 3Y$ and Scale 2 gives $2Y = 3Z$:

1. From the first relation, $X : Y = 3 : 2$.
2. From the second relation, $Y : Z = 3 : 2$.
3. Bridge the shared variable $Y$ using $\text{LCM}(2, 3) = 6$: set $Y = 6$.
4. Then $2X = 3(6) = 18 \implies X = 9$, and $3Z = 2(6) = 12 \implies Z = 4$.
5. Now every pan is a simple integer sum using $(X, Y, Z) = (9, 6, 4)$.

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2. Lever Torque & Hanging Mobile Balances (Eşit Bölmeli Çubuk Dengesi)

An independent lever exercise replaces pans with a horizontal notched beam (eşit bölmeli ağırlıksız çubuk) suspended from a ceiling cord or resting on a knife-edge fulcrum (destek noktası). For the examples here the rod is explicitly weightless. If a question gives rod mass, include its weight acting at the centre of mass.

Rotational Torque Equilibrium (Moment Dengesi)

For any horizontal beam to remain balanced around its support point, the total counterclockwise torque on the left must equal the total clockwise torque on the right:

$$\sum W_{\text{left}} \cdot d_{\text{left}} = \sum W_{\text{right}} \cdot d_{\text{right}}$$

where $d_i$ is the number of equal grid intervals (bölme sayısı) measured horizontally from the support fulcrum to the hanging weight $W_i$.

Multi-Tier Hanging Mobiles (Asılı Makara ve Çubuk Sistemleri)

When a lower horizontal bar hangs by a vertical string from an upper bar, apply two laws from the bottom tier upward:

1. Torque Law on Each Bar: Balance $\sum W_{\text{left}} d_{\text{left}} = \sum W_{\text{right}} d_{\text{right}}$ around that bar's own suspension string.
2. Total Downward Pull on the Suspension String: The upward tension in the vertical string holding a weightless bar equals the sum of all weights hanging from that bar ($T_{\text{string}} = W_1 + W_2 + \dots$). That total weight then acts at a single attachment notch on the bar above it.

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3. Tilted / Inequality Scales (Dengede Olmayan Teraziler)

When the scale beam is tilted (eğik), the pans are not equal:

• Lower (Sunk) Pan (Aşağıdaki Kefe): Contains the heavier total weight ($>$).
• Higher (Raised) Pan (Yukarıdaki Kefe): Contains the lighter total weight ($<$).

Algebraic Rules for Inequality Scales

1. Cancelation Preserves Inequality Sense: Removing identical weights from both pans keeps the tilt direction unchanged:

$$A + C > B + C \implies A > B$$

1. Transitive Ordering (Geçişme Özelliği): If Scale 1 shows $A > B$ and Scale 2 shows $B > C$, then $A > B > C$.
2. Same-Sense Addition (Aynı Yönlü Eşitsizliklerin Toplanması): If the left pan is lower on both Scale 1 ($L_1 > R_1$) and Scale 2 ($L_2 > R_2$), you can add the two heavier pans and the two lighter pans:

$$L_1 + L_2 > R_1 + R_2$$

Warning: Never subtract two inequalities directly! If you know $A + B > C + D$ and a second scale shows $B < D$ (meaning $B$ is lighter than $D$), rewrite the second inequality in the same sense as $D > B$, add them ($A + B + D > C + D + B$), and cancel $B + D$ from both sides to conclude strictly that $A > C$.

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Summary Table of Balance Scale Laws

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Worked Examples

Worked Example 6.1: Three-Shape Equilibrium via LCM Weight Assignment

Problem: Two balanced scales are shown with circles (●, weight $C$), squares (■, weight $S$), and triangles (▲, weight $T$):

• Scale 1: Left pan holds $2C + S$; right pan holds $2S + T$.
• Scale 2: Left pan holds $S + T$; right pan holds $3T$.

How many triangles ($T$) must be placed on the right pan of a third scale to balance $4C + S$ on its left pan?

Solution:

1. Simplify Scale 2 by opposite-pan cancelation:

$$S + T = 3T \implies S = 2T$$

1. Simplify Scale 1 by canceling one square ($S$) from both pans:

$$2C + S = 2S + T \implies 2C = S + T$$

1. Assign integer weights in terms of $T$: Let $T = 2$. Then from Scale 2:

$$S = 2(2) = 4$$

Substitute into simplified Scale 1:

$$2C = 4 + 2 = 6 \implies C = 3$$

1. Evaluate the target pan ($4C + S$):

$$4C + S = 4(3) + 4 = 16$$

Since each triangle ($T$) has weight $2$, the number of triangles needed is $16 / 2 = 8T$.

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Worked Example 6.2: Deductions on Tilted Inequality Scales

Problem: Three distinct blocks $X, Y, Z$ are tested on two tilted scales:

• Scale 1: Pan $(X + Y)$ is lower than pan $(Y + Z)$.
• Scale 2: Pan $(2Z)$ is lower than pan $(X + Y)$.

Order the weights $X, Y, Z$ from lightest to heaviest.

Solution:

1. Translate Scale 1 (lower pan is heavier):

$$X + Y > Y + Z$$

Cancel $Y$ from both pans:

$$X > Z$$

1. Translate Scale 2 (lower pan is heavier):

$$2Z > X + Y$$

1. Chain the inequalities using $X > Z$: Since $X > Z$, adding $Y$ to both sides gives $X + Y > Z + Y$. Combining this with Scale 2 yields:

$$2Z > X + Y > Z + Y \implies 2Z > Z + Y$$

Subtracting $Z$ from both sides proves $Z > Y$. 4. Combine both strict orderings: Since $Y < Z$ and $Z < X$, the unique order from lightest to heaviest is $Y < Z < X$ (for instance, $(X, Z, Y) = (6, 5, 3)$ satisfies both $6 + 3 > 3 + 5$ and $2(5) > 6 + 3$).

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Worked Example 6.2B: Two-Tier Notched Beam Mobile (İki Katlı Eşit Bölmeli Çubuk)

Problem: A weightless lower rod is divided into $6$ equal unit segments and hangs from an upper rod by a vertical cord attached at notch $2$ (meaning $2$ units from its left end and $4$ units from its right end). A block $P$ hangs from the far-left end of the lower rod, and a block $Q = 15\text{ g}$ hangs from the far-right end. The upper rod is supported at its center fulcrum $O$: the cord holding the lower rod hangs at distance $3$ units to the left of $O$, and a counterweight $R$ hangs at distance $5$ units to the right of $O$. Find the weights of $P$ and $R$ so that both rods remain horizontal.

Solution:

1. Balance the lower rod around its suspension cord: The left arm has length $d_P = 2$ units and the right arm has length $d_Q = 4$ units:

$$P \cdot 2 = Q \cdot 4 \implies 2P = 15 \cdot 4 = 60 \implies P = 30\text{ g}$$

1. Find the total downward tension on the suspension cord: Because the lower rod is weightless, the cord supports the sum of both hanging weights:

$$T_{\text{cord}} = P + Q = 30 + 15 = 45\text{ g}$$

1. Balance the upper rod around fulcrum $O$: The cord pulls down with $45\text{ g}$ at distance $3$ units left of $O$, while $R$ pulls down at distance $5$ units right of $O$:

$$45 \cdot 3 = R \cdot 5 \implies 135 = 5R \implies R = 27\text{ g}$$

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Common TR-YÖS Balance Scale Traps to Avoid

• Trap 1 — Subtracting Same-Sense Inequalities: If $A + B > C + D$ and $A > D$, you cannot subtract the second inequality from the first. Only add inequalities that point in the exact same direction ($>$ with $>$).
• Trap 2 — Forgetting Total Weight on Upper Mobile Strings: When moving from a lower balanced beam to an upper beam, the string carries $W_{\text{left}} + W_{\text{right}}$, not the torque product.
• Trap 3 — Solving for Every Shape Individually: Before isolating $X, Y, Z$ one by one, always check if $(L_1 + L_2)$ or $(2L_1 - L_2)$ immediately forms the question's target pan.
• Trap 4 — Measuring Lever Arms from the Rod End Instead of the Fulcrum: Always count the grid distance $d_i$ from the support fulcrum $O$ or suspension string, never from the left tip of the beam. On a 6-unit rod supported at notch 2, a weight at the right tip has torque arm $6 - 2 = 4$ units (not $6$ units), and a weight at notch 1 has horizontal torque arm $d = 2 - 1 = 1$ unit from $O$.

Review the Method

On tilted inequality scales (dengede olmayan teraziler), the lower (sunk) pan is strictly heavier than the higher (raised) pan; combining transitive inequalities determines the heaviest or lightest shape.