A balanced lever has two weights on it, the first with mass $8 k g$ $8 kg$ and the second with mass $30 k g$ $30 kg$ . If the first weight is $7 m$ $7 m$ 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 $8 k g$ $8 kg$ and the second with mass $30 k g$ $30 kg$ . If the first weight is $7 m$ $7 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2015-12-26T13:34:24

$1.867 m$ $1.867m$

Explanation:

I will apply the second condition of static equilibrium which states that the resultant torque (moments) must be zero,
ie. $\sum τ = 0$ $sum tau=0$
where $\to τ = \to r \times \to F$ $vectau=vecr xx vecF$ is the algebraic cross product of the position vector from the axis of rotation (fulcrum) to the applied force, and the applied force.

So in this particular case,

$r_{1} F_{1} - r_{2} F_{2} = 0$ $r_1F_1-r_2F_2=0$

$therefore(7xx8g)-(r_2xx30g)=0$

$thereforer_2=(7xx8xx9,8)/(30xx9,8)$

$=1.867m$ .

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A balanced lever has two weights on it, the first with mass $8 k g$ $8 kg$ and the second with mass $30 k g$ $30 kg$ . If the first weight is $7 m$ $7 m$ 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 8 kg 8kg8 kg and the second with mass 30 kg30kg30 kg. If the first weight is 7 m7m 7 m from the fulcrum, how far is the second weight from the fulcrum?

1 Answer

Explanation:

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A balanced lever has two weights on it, the first with mass $8 k g$ $8 kg$ and the second with mass $30 k g$ $30 kg$ . If the first weight is $7 m$ $7 m$ from the fulcrum, how far is the second weight from the fulcrum?