A plane is inclined at \(30^{\circ}\) to the horizontal. If a load is dragged up the plane, the ratio of the distance moved by the load to the distance moved by the effort is ____________
Explanation
For an inclined plane,
V. \(R=\frac{1}{\sin \theta}\)
Then, for the inclined plane given,
V. \(R=\frac{1}{\sin 30^{\circ}}=2\)
However, the parameter required is not the velocity ratio but the multiplicative inverse (the reciprocal) of the velocity ratio.
\(V . R=\frac{\text { Distance moved by effort }}{\text { Distance moved by load }}\)
Then,
\(\frac{\text { Distance moved by load }}{\text { Distance moved by effort }}=\frac{1}{V \cdot R}=\frac{1}{2}\).
Hence, the required ratio is \(1: 2\) (not \(2: 1\) )
