A student investigates the balancing of a metre rule pivoted at the 50.0 cm50.0\text{ cm}50.0 cm mark.
An object QQQ is placed at the 85.0 cm85.0\text{ cm}85.0 cm mark on the rule. A load of variable weight PPP is placed on the other side of the pivot at a distance xxx from the pivot to keep the metre rule balanced and horizontal.
The table below shows the results of the measurements:
| P/NP / \text{N}P/N | x/cmx / \text{cm}x/cm | 1x/cm−1\frac{1}{x} / \text{cm}^{-1}x1/cm−1 |
|---|---|---|
| 1.51.51.5 | 35.035.035.0 | 0.02860.02860.0286 |
| 2.02.02.0 | 26.326.326.3 | 0.03800.03800.0380 |
| 3.03.03.0 | 17.517.517.5 | 0.05710.05710.0571 |
| 4.04.04.0 | 13.113.113.1 | (i) |
| 5.05.05.0 | (ii) | 0.09520.09520.0952 |
Calculate the missing values (i) and (ii) to 3 significant figures.
The maximum possible distance of the load from the pivot in this setup is xmax=50.0 cmx_{\text{max}} = 50.0\text{ cm}xmax=50.0 cm. Calculate the value of 1xmax\frac{1}{x_{\text{max}}}xmax1.
The relationship between PPP and xxx is given by:
P=k(1x) P = k \left(\frac{1}{x}\right) P=k(x1)where kkk is a constant. Use the data for P=3.0 NP = 3.0\text{ N}P=3.0 N to calculate the value of kkk in N⋅cm\text{N}\cdot\text{cm}N⋅cm.
Use your previous results to determine the minimum weight PPP required to balance the metre rule in this experiment.
245 exam-style questions on CIE IGCSE Physics 1.5 Forces, covering 1.5.1a Forces, resultant force and motion, 1.5.1b Load–extension graphs and the spring constant, 1.5.1c Friction and drag, 1.5.1d Force, mass and acceleration, 1.5.1e Motion in a circular path, 1.5.2a Moment of a force, 1.5.2b Principle of moments and equilibrium, and 1.5.3 Centre of gravity and stability. Each one has a worked solution and a mark scheme showing where the marks go.