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A balanced lever has two weights on it, the first with mass $9 k g$ $9 kg$ and the second with mass $32 k g$ $32 kg$ . If the first weight is $6 m$ $6 m$ from the fulcrum, how far is the second weight from the fulcrum?

2 Answers

Feb 25, 2016

$\frac{27}{16}$ $27/16$ m.

Explanation:

For a balanced lever the torque on the two ends must be equal.

Torque can be defined as $mass \times distance from fulcrum$ $"mass" xx "distance from fulcrum"$

So
$XXX m_{1} \times d_{1} = m_{2} \times d_{2}$ $color(white)("XXX")m_1xxd_1 = m_2 xx d_2$

or for the given data:
$XXX 9 kg \times 6 m = 32 kg \times d_{2} m$ $color(white)("XXX")9 " kg"xx6" m" = 32" kg" xx d_2 " m"$

$XXX d_{2} = \frac{9 \times 6}{32} m = \frac{27}{16} m$ $color(white)("XXX")d_2=(9xx6)/32 " m" = 27/16" m"$

Shwetank Mauria

Feb 25, 2016

Second weight is $\frac{54}{32} = 1.6875$ $54/32=1.6875$ $m$ $m$ far from fulcrum.

Explanation:

Let the the second weight be $x$ $x$ m far from the fulcrum.

As the lever is balanced, moment ( $M = F \cdot d$ $M=F*d$ , where $F$ $F$ is force applied and $d$ $d$ is its distance from fulcrum) of the two weights should be equal.

Moment of first weight is $9 \cdot 6 = 54$ $9*6=54$ and as it balances second weight of $32$ $32$ kg, the latter's distance should be

$\frac{54}{32} = 1.6875$ $54/32=1.6875$ $m$ $m$ .

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