A balanced lever has two weights on it, one with mass $3 k g$ $3 kg$ and one with mass $8 k g$ $8 kg$ . 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 · 2016-08-27T02:53:04

$\frac{21}{8}$ $21/8$ The force times distance on one side must balance the force times distance on the other side.

$M_{1} \times D_{1} = M_{2} \times D_{2}$ $M_1 xx D_1 = M_2 xx D_2$

The force x distance on one side of the fulcrum must equal the force times the distance on the other side of the fulcrum.

$M_{1}$ $M_1$ = 3 kg

$D_{1}$ $D_1$ = 7 m

$M_{2}$ $M_2$ = 8kg

$D_{2}$ $D_2$ = an unkown.

Subsitute the values into the equation and solve.

$3 \times 7 = 8 \times D_{2}$ $3 xx 7 = 8 xx D_2$

$\frac{21}{8} = D_{2}$ $21/8 = D_2$

$2 \frac{5}{8} = D_{2}$ $2 5/8 = D_2$

A balanced lever has two weights on it, one with mass 3 kg3kg3 kg and one with mass 8 kg8kg8 kg. If the first weight is 7 m7m 7 m from the fulcrum, how far is the second weight from the fulcrum?