Question

A pulley system with VR = 2 is used to lift a load of 175 kgf through a vertical height of 15 m. The effort
required is 100 kgf in the downward direction (g=10 Nkg-1). Calculate
(i) Distance moved by the effort
(ii) Work done by the effort
(iii) M.A of the pulley system
(iv) Efficiency of the pulley system​

Answer 1

Answer:

1. Distance moved by effort 30m

2. Work done by effort =3000 J

3. Mechanical advantage of pulley system= 1.75

4. Efficiency of pulley system= 87.5%

Explanation:

1. Displacement due to to effort is VR =2

load (L)= 175 kgf

effort (E)= 100 kgf

Load displacement (dl) = 15m

VR= de/dl =

effort displacement (de)= 2 × dl = 2×15= 30 m

2. work done by by the effort (WE)

= Force (effort) × de

= 100×30

= 3000 J

3. Mechanical advantage of pulley system=$\frac{load \: (l)}{effort(e)}$

=175/100 = 1.75

4. Efficiency of pulley system=

work output/ work input

= Work done by load/ Work done by effort

here, work done by load = force load × dl = 175× 15

= 2625kgf

so, efficiency= 2625/3000

= 0.875× 100

= 87.5 %