You are provided with a wooden block to which a hook is fixed, a set of masses, spring balance, and other necessary materials. Using the diagram above as a guide, carry out the following instructions.
- Record the mass m\(_{0}\), indicated on the wooden block.
- Place the block on the table.
- Attach the spring balance to the hook.
- Pull the spring balance horizontally with a gradual increase in force until the block just starts to move Record the spring balance reading F.
- Repeat the procedure by placing in turn mass m=200, 400, 600, and 800g on top of the block. In each case, read and record the corresponding value of F.
- Evaluate M = m\(_{0}\) + m and R = \(\frac{m}{100}\) in each case
- Tabulate your readings.
- Plot a graph with F on the vertical axis and R on the horizontal axis
- Determine the slope, s, of the graph.
- State two precautions taken to ensure accurate results.
(b)i. Define coefficient of static friction.
ii. A block of wood of mass 0.5 kg is pulled horizontally on a table by a force of 2.5 N. Calculate the coefficient of static friction between the two surfaces.(g = 10ms\(^{-2}\))
Explanation
Observation /table of values
Mass mo = 400g.
| S/N | mo(g) | m(g) | (M=mo+m)g | R = \(\frac{M(g)}{100}\) | F(N) |
| 1 | 400.0 | 0.00 | 400.0 | 4.00 | 1.70 |
| 2 | 400.0 | 200.0 | 600.0 | 6.00 | 2.40 |
| 3 | 400.0 | 400.0 | 800.0 | 8.00 | 3.60 |
| 4 | 400.0 | 600.0 | 1000.0 | 10.00 | 5.20 |
| 5 | 400.0 | 800.0 | 1200.0 | 12.00 | 6.30 |
Slope, s = \(\frac{\bigtriangleup{F}}{\bigtriangleup{R}} = \frac{F_{2}-F_{1}}{R_{2}-R_{1}}\)
Slope (S) = \(\frac{6.4-2.3}{12.0-6.0}\)
S = \(\frac{4.1}{6.0}\) = 0.68
PRECAUTION; i
- Avoided parallax error in reading spring balance.
- Tapped the wood/load slightly to avoid sticking to the table.
- Repeated readings are shown in the table.
- Zero error of the spring balance.
(b)i. Coefficient of static friction: This is the ratio of the frictional force F (when the body is just about to move) to the normal reaction R (between the two surfaces in contact).
ii. \(\mu\) = \(\frac{F}{R} = \frac{2}{5}\) = 0.5