A pulley system pulls a load of 600N by an effort of 200N.If the resistance due to the movable parts of a machine is 400N,find:

1.MA

2.VR

3.No.of pulley

4.Efficiency

Asked by pb_ckt | 4th Feb, 2019, 12:27: PM

Expert Answer:

Load (L) = 600 N 
Resistance due to movable parts = 400 N 
 
Effort (E) = 200 N
 
1.
M e c h a n i c a l space a d v a n t a g e space equals space fraction numerator L o a d space left parenthesis L right parenthesis over denominator E f f o r t space left parenthesis E right parenthesis end fraction space equals space fraction numerator 1000 space N space space over denominator 200 space N space end fraction space equals space 5
3. The mechanical advantage of the pulley system = number of pulleys (n) = 5 
 
2. Velocity ratio (VR) = number of pulleys = 5 
 
4. thus, actual mechanical advantage  (MA)= n - (w/E)  = 5 - (400/200) = 3
 
Thus, Efficiency
 
eta equals fraction numerator M A over denominator V R end fraction cross times 100 space equals space 3 over 5 cross times 100 space equals space 60 percent sign
 

Answered by Shiwani Sawant | 4th Feb, 2019, 04:41: PM