Suppose two children are using a uniform seesaw that is 3.00 m long and has its centerof mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from thepivot. (a) Theoretically, how far from the pivot must the second (18.0 kg) child sit tobalance the seesaw. (b) What is unrea

Complete question is;

Suppose two children are using a uniform seesaw that is 3.00 m long and has its center of mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from the pivot. (l

(a) Calculate where the second 18.0 kg child must sit to balance the seesaw. (b) What is unreasonable about the result?

(c) Which premise is unreasonable, or which premises are inconsistent?

A) The second child must sit 2.33 m from the pivot.

B) The pivot is 1.5m from either end of the seesaw.

Now, if the second child is to sit 2.33m from the pivot, it means he sits outside the seesaw as the only area he has to sit is 1.5m long. Thus, this is unreasonable.

C) The unreasonable premise is the distance from the pivot where the first child sits. If the first child sits much closer to the pivot, the distance at which the 2nd child sits to balance to seesaw will fall on the seesaw.

Explanation:

We are given;

Mass of first child; m1 = 30 kg

Mass of second child; m2 = 18 kg

Length of seesaw; L = 3m

Distance of 1st child from pivot; x1 = 1.4m

Let the distance of the 2nd child from the pivot be x2

A) Let's find the weight of each child.

Weight of 1st child;

W1 = m1 × g

W1 = 30 × 9.81

W1 = 294.3 N

Weight of second child;

W2 = m2 × g

W2 = 18 × 9.81

W2 = 176.57 N

Taking moments about the pivot, we have;

M1 = W1 × x1

M1 = 294.3 × 1.4

M1 = 412.02 N.m

M2 = W2 × x2

M2 = 176.57 × x2

For the seesaw to be balanced, M1 must be equal to M2.

Thus,

412.02 = 176.57 × x2

x2 = 412.02/176.57

x2 = 2.33 m

Thus, the second child must sit 2.33 m from the pivot.

B) We are told that the center of mass is over the pivot. Thus, the pivot is 1.5m from either end of the seesaw.

Now, if the second child is to sit 2.33m from the pivot, it means he sits outside the seesaw as the only area he has to sit is 1.5m long. Thus, this is unreasonable.

C) The unreasonable premise is the distance from the pivot where the first child sits. If the first child sits much closer to the pivot, the distance at which the 2nd child sits to balance to seesaw will fall on the seesaw.

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