You are provided with a uniform metre rule, a knife edge and a body m of mass 50g. Suspend the given body m by means of a thread from 1.0cm mark of the metre rule. Balance the loaded metre rule on the knife-edge as shown in the diagram above. Determine and record the value of x when the metre rule is in horizontal equilibrium. Evaluate \(\frac{1}{x}\). Repeat the experiment for values of m = 70, 90, 110, 130 and 150g respectively. In each case, determine and record the corresponding values of x and \(\frac{1}{x}\). Tabulate your readings. Plot a graph of m on the vertical axis and \(\frac{1}{x}\) on the horizontal axis, starting both axes from the origin (0,0). Determine the slope s, of the graph and the value of m for which /\(\frac{1}{x}\) = 0. State two precautions taken to ensure accurate results.
(b)i. Using your graph, determine the value of x for which m = 0.
(ii) State two conditions necessary to maintain the metre rule in the experiment above, in equilibrium.
(iii) Using your graph, determine the value of x for which m = 100g.
Explanation
Table of values/Observation
| S/N | m(g) | x(cm) | (cm) |
| 1 | 50.00 | 49.75 | 0.0202 |
| 2 | 70.00 | 39.65 | 0.0252 |
| 3 | 90.00 | 35.10 | 0.0285 |
| 4 | 110.00 | 30.75 | 0.0325 |
| 5 | 130.00 | 27.40 | 0.0365 |
| 6 | 150.00 | 25.00 | 0.0400 |
Slope (s) = \(\frac{\bigtriangleup {m(g)}}{\bigtriangleup {\frac {1}{x}(cm^{-1})}}\) = \(\frac{130-10}{0.04-0.0145}\)
\(\frac{140}{0.0255}\) = 4705.88(g cm)
Deduction: When \(\frac{1}{x}\) = 0, m = \(\frac{1}{x}\)66.00g
Precaution:
- l avoided draught by switching off the fans.
- I ensure that the suspended mass did not touch the table.
- I avoided error due to parallax when reading the metre rule
- Zero error of the metre rule was noted.
(b)i. From the graph, when m = 0, \(\frac{1}{x}\) = 0.0125
\(\therefore\) \(\frac{1}{x} = \frac{1}{0.0125}\) = 80 cm
(1) Sum of upward forces = sum of downward forces.
(2) Algebraic sum of all the moments about the pivot is Zero.
ii. when m = 100g. X = 0.0305
\(\therefore\) \(\frac{1}{x} = \frac{1}{0.0305}\) = 32.97cm