A balanced lever has two weights on it, the first with mass $16 k g$ $16 kg$ and the second with mass $36 k g$ $36 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?

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A balanced lever has two weights on it, the first with mass $16 k g$ $16 kg$ and the second with mass $36 k g$ $36 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2016-02-16T20:38:13

$0.889 m$ $0.889m$

Explanation:

From the second condition of static equilibrium, the resultant torque must be zero, ie $\sum \to τ = 0$ $sum vec tau = 0$ where $\to τ = \to r \times \to F$ $vec tau = vec r xx vec F$ .
$therefore r_1F_1-r_2F_2=0$
$therefore (2xx16g)-r_2xx36g=0$
$therefore r_2=0.889m$ .

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A balanced lever has two weights on it, the first with mass $16 k g$ $16 kg$ and the second with mass $36 k g$ $36 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?

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A balanced lever has two weights on it, the first with mass 16 kg 16kg16 kg and the second with mass 36 kg36kg36 kg. If the first weight is 2 m2m 2 m from the fulcrum, how far is the second weight from the fulcrum?

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Explanation:

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A balanced lever has two weights on it, the first with mass $16 k g$ $16 kg$ and the second with mass $36 k g$ $36 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?