Question

A pulley system with velocity ratio 3 is used to lift a load of 60 kgf through a height of 20 m. The force is applied in upward direction and its magnitude is 25 kgf. 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. Total number of pulleys in the fixed and movable block.

(Take g = 10 Nkg⁻¹)

Answer 1

Given: V.R = 3

L = 60 kgf

d_L = 20 m

E = 25 kgf

= 25 × 10

= 250 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) = 3 × 20 = 60 m

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

= 250 N × 60 m

= 15000 J

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

\[= \frac{60\ \mathrm{kgf}}{25\ \mathrm{kgf}}\]

= 2.4

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

\[= \frac{2.4}{3} \times 100\]

= 80%

(e) Total number of pulleys : 2 pulleys (1 in the fixed block and 1 in the movable block.)