A balanced lever has two weights on it, the first with mass $8 k g$ $8 kg$ and the second with mass $3 k g$ $3 kg$ . 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 $8 k g$ $8 kg$ and the second with mass $3 k g$ $3 kg$ . 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 · 2016-01-11T10:15:02

$5.333 m$ $5.333m$

Explanation:

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

So application in this case yields :

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

$therefore (2xx8g)-(r_2xx3g)=0$

$therefore r_2=(2xx8xx9.8)/(3xx9.8)=5.333m$ .

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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 $3 k g$ $3 kg$ . 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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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 3 kg3kg3 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 $8 k g$ $8 kg$ and the second with mass $3 k g$ $3 kg$ . If the first weight is $2 m$ $2 m$ from the fulcrum, how far is the second weight from the fulcrum?