A balanced lever has two weights on it, the first with mass $7 k g$ $7 kg$ and the second with mass $25 k g$ $25 kg$ . If the first weight is $6 m$ $6 m$ 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 $7 k g$ $7 kg$ and the second with mass $25 k g$ $25 kg$ . If the first weight is $6 m$ $6 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2015-12-24T09:50:26

$1, 68 m$ $1,68m$

Explanation:

We may use the second condition for static equilibrium to solve this problem, that is, that the resultant torque of the system must be zero.
ie, $\sum \to τ = 0$ $sum vectau=0$ , where $\to τ = \to r \times \to F$ $vectau = vecr xx vecF$ is the vector cross product of the position vector from the axis of rotation (fulcrum) to the applied force and the applied force.

So in this case,

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

$therefore (6xx7g)-(r_2xx25g)=0$

$therefore r_2=(6xx7xx9,8)/(25xx9,8)=1,68m$ .

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A balanced lever has two weights on it, the first with mass $7 k g$ $7 kg$ and the second with mass $25 k g$ $25 kg$ . If the first weight is $6 m$ $6 m$ 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 7 kg 7kg7 kg and the second with mass 25 kg25kg25 kg. If the first weight is 6 m6m 6 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 $7 k g$ $7 kg$ and the second with mass $25 k g$ $25 kg$ . If the first weight is $6 m$ $6 m$ from the fulcrum, how far is the second weight from the fulcrum?