Question

A pulley system with VR = 4 is used to lift a load of 175 kgf through a vertical height of 15 m. The effort required is 50 kgf in the downward direction. (g = 10 N kg^-1)

Calculate:

1. Distance moved by the effort
2. Work done by the effort
3. M.A. of the pulley system
4. Efficiency of the pulley system

Answer 1

1) Displacement due to the effort is VR = 4

Load (L) = 175 kgf
Effort (E) = 50 kgf

Load displacement (d_L) = 15 m

`VR = d_E/d_L`

Effort displacement (d_E) =  4 × d_L

= 4 × 15 

= 60 m

Distance moved by the effort is 60 m

2) Work done by the effort is calculated using the following formula.

Work done by effort (W_E) = `"Force"_"Effort" xx d_E`

W_E = 50 × 60 

= 3000 J

3) Mechanical advantage of the pulley system is calculated as

MA of the pulley system = `"Load (L)"/"Effort (E)"`

= `175/50`

= 3.5

4) Efficiency of the pulley is the ratio of work output to the work input required by the system

Efficiency of the pulley system (η) = `"Work output (Work Done by Load)"/"Work input (Work Done by Effort)"`

Work done by load = Force_Load x d_L

= 175 × 15

= 2625 kgf

:. η = `2625/3000`

= 0.875 × 100

= 87.5 %