Question

To balance a seesaw, the distance a person is from the fulcrum is inversely proportional to his or her weight . Roger , who weighs 120 pounds is sitting 6 feet from the fulcrum . Ellen weighs 108 pounds . How far from the fulcrum must she sit to balance the seesaw ? Round to the nearest hundredth of a foot .

Answer 1

Answer:

$d_e =6.67ft$

Step-by-step explanation:

From the question we are told that:

Weight of Roger $W_r=120\ pounds$

Distance of Roger from fulcrum $d_r=6 ft$

Weight of Ellen $W_e=120\ pounds$

Generally the equation for distance-weight relationship is mathematically given by

  $d\alpha \frac{1}{W}$

  $\frac{d_1}{d_2} =\frac{W_2}{W_1}$

  $\frac{d_r}{d_e} =\frac{W_e}{W_r}$

Therefore

 $\frac{d_e}{d_r} =\frac{W_r}{W_e}$

 $d_e =\frac{W_r*d_r}{W_e}$

 $d_e =\frac{6*120}{108}$

 $d_e =6.67ft$

Therefore the distance from the fulcrum she must sit to balance the seesaw is given as

$d_e =6.67ft$