You are provided with a uniform metre rule, a knife-edge, some masses and other necessary materials. i. Determine and record the centre of gravity of the me...
Question 1 Report
You are provided with a uniform metre rule, a knife-edge, some masses and other necessary materials.
i. Determine and record the centre of gravity of the metre rule.
ii. Fix the 100g mass marked N at a point Y, the 80cm mark of the rule using a sellotape.
iii. Suspend another 50g mass marked M at X, a distance A = 1Ocm from the 0cm mark of the rule.
iv. Balance the arrangement horizontally on the knife edge as illustrated in the diagram above.
v. Measure and record the distance B of a knife-edge from the 0cm mark of the rule.
vi. Repeat the procedure for four other values of A =15cm, 20cm, 25cm and 30cm.
vii. Tabulate your readings.
viii. Plot a graph with B on the vertical axis and A on the horizontal axis.
ix. Determine the slope, s, of the graph.
x. Also determine the intercept, c, on the vertical axis.
xi. Evaluate:
\(\propto\)) = k\(_{1}\) = (\(\frac{1 - 2s}{s}\))100
(\(\beta\)) = k\(_{2}\) = \(\frac{2c}{s}\) = 160
xii. State two precautions taken to obtain accurate results.
(b)i. Define the moment of a force about a point.
ii. State two conditions under which a rigid body at rest remains in equilibrium when acted upon by non-parallel coplanar forces.
Answer Details
(i) Centre of gravity of the uniform metre rule = 50.0 cm mark (the rule balances horizontally at this point before any mass is added).
The 100 g mass N is fixed with sellotape at the 80 cm mark (point Y); the 50 g mass M is suspended at point X, a distance A from the 0 cm end. The knife-edge is moved until the rule balances horizontally, and B, the distance of the knife-edge from the 0 cm end, is recorded.
(vii) Table of readings
| S/N | A / cm | B / cm |
| 1 | 10.0 | 54.0 |
| 2 | 15.0 | 55.0 |
| 3 | 20.0 | 56.0 |
| 4 | 25.0 | 57.0 |
| 5 | 30.0 | 57.0 |
| 6 | 35.0 | 58.0 |
(viii) Graph of B against A
B is plotted on the vertical axis against A on the horizontal axis and the best straight line is drawn through the points.
(ix) Slope, s
Using two widely-spaced points on the best line, \((A_1,B_1)=(0,\,53.0)\) and \((A_2,B_2)=(30.0,\,59.0)\):
\[s=\frac{B_2-B_1}{A_2-A_1}=\frac{59.0-53.0}{30.0-0}=\frac{6.0}{30.0}=0.2\](x) Intercept, c, on the vertical axis (where the line cuts A = 0):
\[c=53.0\ \text{cm}\](xi) Evaluation
\[k_1=\left(\frac{1-2s}{s}\right)100=\left(\frac{1-2(0.2)}{0.2}\right)100=\left(\frac{1-0.4}{0.2}\right)100=\left(\frac{0.6}{0.2}\right)100=300\]\[k_2=\frac{2c}{s}-160=\frac{2(53.0)}{0.2}-160=\frac{106.0}{0.2}-160=530-160=370\](xii) Two precautions
- The rule was made to balance exactly horizontal on the knife-edge before each reading of B was taken.
- Error due to parallax was avoided by reading the metre rule with the eye placed vertically above the mark being read.
(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 the point to the line of action of the force. Its unit is the newton-metre (N m).
(b)(ii) Two conditions for equilibrium under non-parallel coplanar forces
- The vector sum of all the forces is zero (i.e. the algebraic sum of their resolved components in any given direction is zero).
- The algebraic sum of the moments of the forces about any point is zero.
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