Question

A pulley system with velocity ratio 4 is used to lift a load of 100 kgf through a vertical height of 15 m. The effort required to do so is 40 kgf which is applied in the downward direction. Calculate:

1. Distance through which effort is applied
2. Work done by the effort
3. Mechanical advantage of pulley system
4. Efficiency of pulley system
5. Number of pulleys in the upper and lower block.

(Take g = 10 Nkg⁻¹)

Answer 1

Given: V.R = 4

L = 100 kgf

d_L = 15 m

E = 40 kgf

= 40 × 10

= 400 N

d_E = ?

\[ \text{(a) V.R} = \frac{\text{distance through which effort moves}}{\text{distance through which load moves}} = \frac{d_E}{d_l} \]

∴ Distance through which effort moves (d_E) = 4 × 15 = 60 m

(b) Work done by the effort = Effort × d_E

= 400 N × 60 m

= 24000 J

\[ \text{(c) Mechanical advantage} = \frac{L}{E}\]

\[= \frac{100\ \mathrm{kgf}}{40\ \mathrm{kgf}}\]

= 2.5

\[ \text{(d) Efficiency } (\eta) = \frac{\mathrm{MA}}{\mathrm{VR}} \times 100\]

\[= \frac{2.5}{4} \times 100\]

= 62.5%

(e) Total number of pulleys : 4 (2 in upper fixed block and 2 in lower movable block.)