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 ...
Question 1 Report
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\(^{-2}\))
Answer Details
(a) Static friction experiment
The block just begins to move when the horizontal pull \(F\) equals the limiting (static) friction. The limiting friction is proportional to the normal reaction, and the normal reaction equals the total weight \(M = m_0 + m\). The mass marked on the wooden block is \(m_0 = 400\,\text{g}\).
Observation / table of values (\(m_0 = 400.0\,\text{g}\)):
| S/N | \(m_0\) (g) | \(m\) (g) | \(M = m_0+m\) (g) | \(R = \dfrac{M}{100}\) | \(F\) (N) |
|---|---|---|---|---|---|
| 1 | 400.0 | 0.0 | 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 |
Graph of \(F\) against \(R\)
Slope of the graph
Taking two well-separated points on the line of best fit, \((R_1, F_1) = (6.0,\ 2.3)\) and \((R_2, F_2) = (12.0,\ 6.4)\):
\[ s = \frac{F_2 - F_1}{R_2 - R_1} = \frac{6.4 - 2.3}{12.0 - 6.0} = \frac{4.1}{6.0} \]\[ s = 0.68 \]Two precautions
- I avoided parallax error by reading the spring balance with my eye directly in line with the pointer.
- I ensured the spring balance was pulled strictly horizontally so that the full weight acted as the normal reaction, and I noted the reading at the exact instant the block just started to move.
(b)(i) Definition
The coefficient of static friction is the ratio of the limiting (maximum) frictional force \(F\), acting just as the body is about to move, to the normal reaction \(R\) between the two surfaces in contact:
\[ \mu_s = \frac{F}{R} \](b)(ii) Calculation
Normal reaction:
\[ R = mg = 0.5 \times 10 = 5\,\text{N} \]At the point of moving, the limiting friction equals the applied force, \(F = 2.5\,\text{N}\).
\[ \mu_s = \frac{F}{R} = \frac{2.5}{5} = 0.5 \]The coefficient of static friction between the two surfaces is 0.5.
Download The App On Google Playstore
Everything you need to excel in your exams
