# Civil Engineering - Applied Mechanics - Discussion

Discussion Forum : Applied Mechanics - Section 1 (Q.No. 41)
41.
A load of 500 kg was lifted through a distance of 13 cm. by an effort of 25 kg which moved through a distance of 650 cm. The efficiency of the lifting machine is
50%
40%
55%
30%.
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

Siddhi said:   2 years ago

MA = W/P = 500/25 = 20.
Velocity ratio (VR) = (Distance moved by the effort (Y))/(Distance moved by the load (X)).
VR = Y/X = 650/13 = 50.
Efficiency = MA/VR = 20/50 = 0.4 i.e. 40%.

Option B is the right.

Siddhi study point said:   2 years ago

Hence option B is right.

Naafay Achakzai said:   3 years ago
Weldone, Thanks @Abhay.

Hitman said:   6 years ago
Most perfect and easy solution, Thanks @Nikil.

Nikil said:   7 years ago
Efficiency = (output)/(input)
Input = effort*distance through which the efforts act.

Efficency = (500*13)/(25*650) = 0.4.
Hence efficiency in % would be 40%.

Abhay said:   8 years ago

MA = W/P = 500/25 = 20.

Velocity ratio (VR) = (Distance moved by the effort (Y))/(Distance moved by the load (X)).

VR = Y/X = 650/13 = 50.

Efficiency = MA/VR = 20/50 = 0.4 i.e. 40%.

Niraj kareliya said:   8 years ago
Efficient is equals to MA/VR and MA is equals to W/P and VR is equaled to Y/X.

Harshit said:   10 years ago
Efficiency = m.a/w.p*100.
m.a = w/p = 500/25 = 20.

v.r = y/x = 650/13 = 50.
N = 20/50*100 = 40%.