Fix the 100g mass marked P at B, the 80 cm mark of the uniform metre rule, using an adhesive. Suspend another 100g mass marked Q at A, a distance V = 10.Ocm...
Question 1 Report
- Fix the 100g mass marked P at B, the 80 cm mark of the uniform metre rule, using an adhesive.
- Suspend another 100g mass marked Q at A, a distance V = 10.Ocm from the 0 cm mark of the metre rule.
- Balance the whole arrangement horizontally on a knife edge as shown in the diagram above.
- Measure and record the distance U of K from the 0 cm mark of the metre rule.
- Repeat the procedure for five other values of V = 15.0, 20.0, 25.0, 30.0 and 35.0 cm
- In each case, measure and record the Corresponding values of U. Tabulate your readings.
- Plot a graph of U on the vertical axis against V on the horizontal axis.
- determine the:
(1) slope, s, of the graph
(2) intercept c, on the vertical axis. - Evaluate (i) \(K_{1} = \frac{(1 - 2s)}{s} 100\):
(ii) \(K_{2} = \frac{2c}{s} - 160\) - State two precautions taken to ensure an accurate result
(b)i. State two conditions under which a rigid body at rest remains in equilibrium when acted upon by three non-parallel coplanar forces.
ii. Explain how the position of the centre or gravity of a body affects the equilibrium of the body.
Answer Details
Test of practical knowledge: balancing a metre rule (principle of moments)
The mass P = 100 g is fixed with adhesive at B, the 80.0 cm mark. The second mass Q = 100 g is suspended at A, a distance V from the 0 cm mark. For each value of V the knife edge K is moved until the rule balances horizontally, and its distance U from the 0 cm mark is read.
Table of readings
| V / cm | U / cm |
|---|---|
| 10.0 | 47.0 |
| 15.0 | 48.3 |
| 20.0 | 49.4 |
| 25.0 | 51.2 |
| 30.0 | 52.5 |
| 35.0 | 54.0 |
Graph of U against V
Plotting U on the vertical axis against V on the horizontal axis gives a straight line of best fit through the points.
(1) Slope of the graph
Taking two widely separated points on the line of best fit, \((V_1,U_1)=(10.0,\;47.0)\) and \((V_2,U_2)=(35.0,\;54.0)\):
\[s=\frac{\Delta U}{\Delta V}=\frac{54.0-47.0}{35.0-10.0}=\frac{7.0}{25.0}=0.28.\](2) Intercept on the vertical axis
Extending the line of best fit back to \(V=0\) gives
\[c=44\;\text{cm}.\]Evaluation
\[K_{1}=\frac{(1-2s)}{s}\times100=\frac{1-(2\times0.28)}{0.28}\times100=\frac{1-0.56}{0.28}\times100=\frac{0.44}{0.28}\times100=157.\]\[K_{2}=\frac{2c}{s}-160=\frac{2\times44}{0.28}-160=\frac{88}{0.28}-160=314.3-160=154.3.\]Two precautions
- Errors due to parallax were avoided by reading the position of the knife edge on the metre rule with the line of sight directly above the mark.
- Each suspended mass was checked to hang freely without touching the table, and the arrangement was shielded from draught so that it settled to a true balance.
(b)(i) Conditions for equilibrium under three non-parallel coplanar forces
- The three forces must be concurrent - their lines of action must all pass through a single point.
- The resultant (vector sum) of the three forces must be zero, so that they can be represented in magnitude and direction by the three sides of a triangle taken in order.
(b)(ii) Effect of the position of the centre of gravity on equilibrium
The position of the centre of gravity determines the type of equilibrium of a body. When the centre of gravity is low and the vertical line through it falls within the base of support, the body is in stable equilibrium: after a small tilt the weight provides a restoring moment that returns it to its original position. When the centre of gravity is high so that a small tilt makes the vertical line through it fall outside the base, the body is in unstable equilibrium and topples further. When the centre of gravity remains at the same height as the body is displaced, it is in neutral equilibrium. Hence lowering the centre of gravity and widening the base of support both increase the stability of a body.
Download The App On Google Playstore
Everything you need to excel in your exams
