You are designing a block and tackle system for a construction site with following data set: load to be lifted = 3000 N. Maximum effort to be applied = 400 N and efficiency of system = 75%. Calculate :

(a) The minimum V.R. required.

(b) The minimum number of pulleys required.

(c) Suggest a practical pulley arrangement.

Given,

• Load = 3000 N
• Effort = 400 N
• Efficiency = 75% = 0.75

Mechanical advantage (M.A.) of a machine is given by,

$\text {M.A.} = \dfrac{\text {Load}}{\text {Effort}} \\[1em] = \dfrac{3000}{400} \\[1em] = \dfrac{30}{4} \\[1em] = 7.5$

Now, efficiency of a machine is given by,

$\text {Efficiency} = \dfrac {\text {Mechanical advantage (M.A.)}}{\text {Velocity ratio (V.R.)}} \\[1em] \Rightarrow \text {V.R.} = \dfrac{{\text {M.A.}}}{\text {Efficiency}} \\[1em] = \dfrac{7.5}{0.75} \\[1em] = 10$

Hence, the minimum V.R. required is 10.

(b) In a block and tackle system, the velocity ratio (V.R.) is equal to the number of strands of rope supporting the moving block, which is effectively equal to the total number of pulleys used in the system. Since the required V.R. is 10, the system must provide 10 supporting strands.

Therefore,

Number of pulleys = V.R. = 10

Hence, the minimum number of pulleys required is 10.

(c) A practical arrangement to obtain this velocity ratio is to use a block and tackle system with two blocks, one fixed and one movable. The pulleys can be arranged as 5 pulleys in the upper (fixed) block and 5 pulleys in the lower (movable) block. The rope is passed alternately between the two blocks so that the load is supported by 10 segments of the rope, giving the required velocity ratio of 10.