| # | $\theta$ (°) | $\sin\theta$ | $W_1$ (↑) | $W_2$ (↓) | $F = \frac{W_1+W_2}{2}$ | $W\sin\theta$ | Error % |
|---|---|---|---|---|---|---|---|
| No records logged yet. Balance the roller at different angles and click '+ Log Entry'. | |||||||
| # | $\theta$ (°) | $\sin\theta$ | $W_1$ (↑) | $W_2$ (↓) | $F = \frac{W_1+W_2}{2}$ | $W\sin\theta$ | Error % |
|---|---|---|---|---|---|---|---|
| No records logged yet. Balance the roller at different angles and click '+ Log Entry'. | |||||||
1. Principle: Consider a heavy cylindrical roller of weight $W = M g$ resting on a smooth inclined plane of inclination angle $\theta$ with the horizontal. The weight $W$ acts vertically downwards through its center of gravity.
2. Resolution of Forces: Resolving the weight $W$ into two mutually perpendicular components:
3. Eliminating Pulley Friction via Reversible Limits: Let $f$ be the small frictional force of the roller and pulley:
4. $F\text{–}\sin\theta$ Graph Analysis: A graph plotted between the downward force $F$ on the y-axis and $\sin\theta$ on the x-axis is a straight line passing through the origin. Its slope represents the total weight of the roller:
Comparing this slope with the measured weight $W$ on a beam balance verifies the relationship.