Fig. 26.1 shows a seesaw balanced on a pivot at its centre. Three children sit on the seesaw as shown. Child A (400 N) sits 1.5 m from the pivot on the left...
Question 1 Report
Fig. 26.1 shows a seesaw balanced on a pivot at its centre. Three children sit on the seesaw as shown.
Child A (400 N) sits 1.5 m from the pivot on the left. Child B (300 N) sits 0.50 m from the pivot on the left. Child C (weight W) sits 1.2 m from the pivot on the right.
(a) Calculate the total anticlockwise moment about the pivot. [2]
(b) The seesaw is balanced. Calculate W. [2]
(c) Child B moves to a position 1.0 m from the pivot. State whether child C needs to move closer to or further from the pivot to keep the seesaw balanced. [1]
(d) Calculate the new position of child C. [2]
Answer Details
(a) Both children A and B sit on the left side of the pivot, so both contribute anticlockwise moments. Using moment = force x perpendicular distance:
\( \text{Moment from A} = 400 \times 1.5 = 600 \text{ N m} \)
\( \text{Moment from B} = 300 \times 0.50 = 150 \text{ N m} \)
Total anticlockwise moment = 600 + 150 = 750 N m [2]
(b) For the seesaw to be balanced, the principle of moments requires the clockwise moment (from child C on the right) to equal the total anticlockwise moment:
\( W \times 1.2 = 750 \)
\( W = \frac{750}{1.2} = \textbf{625 N} \) [2]
(c) When child B moves from 0.50 m to 1.0 m from the pivot, the anticlockwise moment increases (B's moment doubles from 150 to 300 N m). To restore balance, the clockwise moment must also increase. Since W is fixed at 625 N, child C must move further from the pivot to increase the distance and hence the clockwise moment. [1]
(d) New total anticlockwise moment:
\( (400 \times 1.5) + (300 \times 1.0) = 600 + 300 = 900 \text{ N m} \)
For balance: \( 625 \times d = 900 \)
\( d = \frac{900}{625} = \textbf{1.44 m} \) [2]
This confirms part (c): child C moves from 1.2 m to 1.44 m, i.e. further from the pivot.
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