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A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw? Assume that both people are small enough compared to the length of the seesaw to be thought of as point masses.a))) The side the boy is sitting on will tilt downward.b))) It is impossible to say without knowing the masses.c))) The side the girl is sitting on will tilt downward.d))) It is impossible to say without knowing the distances.e))) Nothing will happen; the seesaw will still be balanced

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Answer:

The system is still balanced

Step-by-step explanation:

If we suppose that the boy weights M and the girl m, and are balanced at distances L1 and L2 from the pivot point respectively, thy will be balanced if the resultant torque of all the farces from the pivot pint is cero:

(1)
T=M*L1-m*L2=0

now they moved to distances (L1)/2 and (L2)/2, the resultant torque will be:

(2)
T=M*(L1)/2-m*(L2)/2

1/2 can be taken out as a common factor:

(3)
T=(M*L1-m*L2)*1/2

As the the inside of the parenthesis equals equation (1) that equals zero, hte whole equiation equals zero:

(3)
T=(0)*1/2=0

So the system is still balanced

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