The maximum safe load L supported by a horizontal wooden beam varies directly as its width… — JAMB UTME

The maximum safe load $L$ supported by a horizontal wooden beam varies directly as its width $w$ and the square of its depth $d$ , and inversely as its length $l$ . A beam of width $6\text{ cm}$ , depth $10\text{ cm}$ , and length $4\text{ m}$ can support a maximum safe load of $900\text{ kg}$ . What is the maximum safe load that can be supported by a beam of the same material having a width of $4\text{ cm}$ , depth $12\text{ cm}$ , and length $6\text{ m}$ ?

$576\text{ kg}$ Answer
B
$480\text{ kg}$
C
$1,296\text{ kg}$
D
$864\text{ kg}$

Answer

576\text{ kg}

The relationship is modeled by

L = \frac{k w d^2}{l}

. Using the initial parameters (

w=6

d=10

l=4

L=900

), we find

k = 6

. Substituting

w=4

d=12

, and

l=6

into the equation gives

L = \frac{6 \times 4 \times 144}{6} = 576\text{ kg}

Step-by-Step Solution

Formulate the variation equation

L = \frac{k \cdot w \cdot d^2}{l}

Direct variation means multiplying factors in the numerator, while inverse variation places the variable in the denominator.

Calculate the constant of variation

k

using initial conditions

900 = \frac{k \cdot 6 \cdot 10^2}{4} \implies 900 = \frac{600 k}{4} = 150 k \implies k = 6

Substitute

L = 900

w = 6

d = 10

, and

l = 4

to solve for

k

Calculate the new load

L

for the new dimensions

L = \frac{6 \cdot 4 \cdot 12^2}{6} = \frac{6 \cdot 4 \cdot 144}{6} = 576\text{ kg}

Substitute

k = 6

w = 4

d = 12

, and

l = 6

into the variation formula.

Key Concept

Joint and Inverse Variation

Estimated Time:1m 30s

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