Question

A pulley system with velocity ratio 6 is used to lift a load 180 kgf through a vertical height of 30 m. The effort required to do so is 50 kgf which is applied in downward direction. Calculate:

3. Distance through which effort is applied
4. Work done by the effort
5. Mechanical advantage of pulley system
6. Efficiency of the pulley system
7. Total number of pulleys in movable and fixed block.

(Take g = 10 Nkg⁻¹)

Answer 1

Given: V.R = 6

L = 180 kgf

d_L = 30 m

E = 50 kgf

= 50 × 10

= 500 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) = 6 × 30 = 180 m

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

= 500 N × 180 m

= 90000 J

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

\[= \frac{180\ \mathrm{kgf}}{50\ \mathrm{kgf}}\]

= 3.6

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

\[= \frac{3.6}{6} \times 100\]

= 60%

(e) Total number of pulleys : 6 (3 in movable and 3 in fixed block.)