(a) You are provided with a metre rule, a knife-edge, set of masses, inextensible string, retort support and other necessary apparatus. i. Place the metre r...
Question 1 Report
(a)
You are provided with a metre rule, a knife-edge, set of masses, inextensible string, retort support and other necessary apparatus.
i. Place the metre rule on the knife edge. Read and record the point G where the metre rule balances horizontally, as shown in Fig (a).
ii. Suspend the metre rule at G with the aid of the string provided and attach the string to the retort support as shown in Fig 1(b). Keep the string attached to this point throughout the experiment.
iii. Attach the mass \(M_{0}\) at the 80cm mark of the metre rule. Determine the distance of y from G. Keep \(M_{0}\) at this position throughout the experiment.
iv. Suspend a mass \(M = 40g\) on the side AG and adjust its position until the metre rule balances horizontally.
v. Measure and record the distance of x of M from G. Evaluate \(x^{-1}\)
vi. Repeat the procedure for four other values of \(M = 60g\), \(80g\), \(100g\) and \(120g\). Measure and record x and evaluate \(x^{-1}\) in each case.
vii. Tabulate the readings.
viii. Plot a graph of M on the vertical axis and x on the horizontal axis, starting both axes from the origin (0,0).
ix. Determine the slope s of the graph:
x. Given that \(s = yM_{0}\), determine \(M_{0}\).
xi. State two precautions taken to obtain accurate results.
(b) i. Define the moment of a force about a point.
ii. A uniform metre rule is suspended by an inextensible string at its centre of gravity. If a mass of 60g is placed at the 25cm mark, what mass should be placed at the 80cm mark of the metre rule to balance it horizontally?
Answer Details
(a) Readings and graph
The balance point of the metre rule is:
\[G=50.0\ \text{cm}\]
The fixed mass \(M_0\) is at the 80.0 cm mark; hence
\[y=80.0-50.0=30.0\ \text{cm}.\]
| S/N | \(M\) (g) | \(x\) (cm) | \(x^{-1}\) (cm\(^{-1}\)) | \(x^{-1}\) \(\times10^{-3}\) cm\(^{-1}\) |
|---|---|---|---|---|
| 1 | 40.0 | 37.50 | 0.027 | 27.00 |
| 2 | 60.0 | 25.00 | 0.040 | 40.00 |
| 3 | 80.0 | 18.75 | 0.053 | 53.00 |
| 4 | 100.0 | 15.00 | 0.067 | 67.00 |
| 5 | 120.0 | 12.50 | 0.080 | 80.00 |
Plot \(M\) on the vertical axis against \(x^{-1}\) on the horizontal axis, with both axes beginning at the origin.
Using two widely separated points on the best-fit line, \((27,40)\) and \((80,120)\):
\[\text{slope}=\frac{120-40}{80-27}=\frac{80}{53}=1.509\ \text{g per }(10^{-3}\text{ cm}^{-1}).\]
Therefore, in base units,
\[s=1.509\times10^3\ \text{g cm}=1509\ \text{g cm}.\]
For equilibrium, taking moments about \(G\),
\[Mx=M_0y,\qquad M=(M_0y)x^{-1}.\]
Thus \(s=M_0y\), and
\[M_0=\frac{s}{y}=\frac{1509}{30.0}=50.3\ \text{g}.\]
Precautions
- The metre-rule positions were read with the eye vertically above the scale to avoid parallax error.
- The rule was allowed to come to rest and balance horizontally before each reading was taken.
(b)(i) The moment of a force about a point is the product of the force and the perpendicular distance between the point and the line of action of the force.
(b)(ii) The centre of gravity is at the 50 cm mark. The 60 g mass is 25 cm from the point of suspension, while the required mass \(m\) at the 80 cm mark is 30 cm from it.
\[m(30)=60(25)\]
\[m=\frac{60\times25}{30}=50\ \text{g}.\]
Hence, the mass required at the 80 cm mark is 50 g.
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