WAEC SSCE - Physics - 2004 (Practical)
Question 1 Report
Using the diagram above as a guide, carry out the following instructions.
- Place the meter rule provided on the knife edge and adjust the position until it balances horizontally.
- Read and record the balance point, G. Keep the knife edge at this point throughout the experiment.
- Suspend a mass Q= 50.0g at a point P 30cm from the 0cm end of the rule.
- On the other side of G, suspend the mass M =30g. Adjust its position until the rule settles down horizontally as shown in the diagram above.
- Read and record the position R of M.
- Record the distance, d, between G and R. Also read and record the distance, a, between P and G.
- Repeat the procedure for four other values of M = 40, 50, 60, and 70 with Q kept in the same position. Evaluate d\(^{-1}\) in each case. Tabulate your readings.
- Plot a graph of M on the vertical axis against d\(^{-1}\) on the horizontal axis.
- Determine the slope,s, of the graph.
- Evaluate k = \(\frac{s}{Q}\)
- State two precautions taken to ensure accurate results.
(b)i. Explain the moment of a force about a point
ii. State the conditions necessary for a body to be in equilibrium when acted upon by a number of parallel end forces
Answer Details
Principle of the experiment
When the metre rule rests horizontally on the knife edge at the balance point \(G\), it is in equilibrium under the two suspended weights. Taking moments about \(G\), the anticlockwise moment of \(Q\) equals the clockwise moment of \(M\):
\[ Q \times a = M \times d \]where \(a\) is the fixed distance \(PG\) and \(d\) is the distance \(GR\). Rearranging for \(M\):
\[ M = (Q\,a)\,d^{-1} \]So a graph of \(M\) (vertical axis) against \(d^{-1}\) (horizontal axis) is a straight line through the origin whose slope is \(s = Q\,a\).
Recorded readings
Balance point: \(G = 49.50\ \text{cm}\). Mass \(Q = 50.0\ \text{g}\) suspended at \(P = 30.0\ \text{cm}\), so the fixed distance
\[ a = PG = 49.50 - 30.00 = 19.50\ \text{cm}. \]For each mass \(M\) the position \(R\) is read, and \(d = R - G\), then \(d^{-1}\) is evaluated.
| S/N | M (g) | R (cm) | d = R − G (cm) | d−1 (cm−1) | d−1 ×10−2 (cm−1) | a (cm) |
|---|---|---|---|---|---|---|
| 1 | 30.0 | 82.00 | 32.50 | 0.0308 | 3.08 | 19.50 |
| 2 | 40.0 | 73.88 | 24.38 | 0.0410 | 4.10 | 19.50 |
| 3 | 50.0 | 69.00 | 19.50 | 0.0513 | 5.13 | 19.50 |
| 4 | 60.0 | 65.75 | 16.25 | 0.0615 | 6.15 | 19.50 |
| 5 | 70.0 | 63.43 | 13.93 | 0.0718 | 7.18 | 19.50 |
Graph of M against d−1
Slope of the graph
Taking two widely separated points on the line of best fit, \((d^{-1}_1,\,M_1) = (0.0308\ \text{cm}^{-1},\ 30.0\ \text{g})\) and \((d^{-1}_2,\,M_2) = (0.0718\ \text{cm}^{-1},\ 70.0\ \text{g})\):
\[ s = \frac{\Delta M}{\Delta (d^{-1})} = \frac{70.0 - 30.0}{0.0718 - 0.0308} = \frac{40.0}{0.0410} = 975\ \text{g cm}. \]Evaluation of k
\[ k = \frac{s}{Q} = \frac{975}{50.0} = 19.5\ \text{g}. \]This equals the fixed distance \(a = PG = 19.50\ \text{cm}\), as expected since \(s = Q\,a\) and therefore \(k = s/Q = a\).
Two precautions
- The scale readings were taken with the eye placed vertically above the mark to avoid the error of parallax.
- The rule was allowed to come completely to rest in the horizontal position before each reading, and the knife edge was kept fixed at \(G\) throughout the experiment.
(b)(i) Moment of a force about a point
The moment of a force about a point is the product of the force and the perpendicular distance from that point (the pivot) to the line of action of the force. Its SI unit is the newton metre (N m).
(b)(ii) Conditions for equilibrium under parallel forces
- The algebraic sum of the forces is zero, i.e. the total upward force equals the total downward force.
- The algebraic sum of the moments of the forces about any point is zero, i.e. the total clockwise moment equals the total anticlockwise moment about that point.
