Mechanical advantage of a lifting machine from load–effort data A 500 kg load is lifted through 13 cm by an effort of 25 kg that moves through 650 cm. Assuming ideal measurement in kgf–cm, what is the mechanical advantage (MA) of the machine?

Introduction / Context:
Mechanical advantage (MA) quantifies how much a machine multiplies force. In simple lifting problems stated in kilogram-force (kgf) units, MA is the ratio of load to effort and is independent of distances. The work distances are used for velocity ratio and efficiency calculations.

Given Data / Assumptions:

• Load = 500 kgf (using kg as kgf as customary in elementary machine problems).
• Effort = 25 kgf.
• Load rise = 13 cm; Effort movement = 650 cm (for context only).
• No slippage assumed in reading MA; efficiency not requested.

Concept / Approach:

Mechanical advantage MA = Load / Effort. The given distances are not needed to compute MA directly; they would relate to the velocity ratio (VR = effort distance / load distance) and efficiency (η = MA/VR), but here only MA is sought.

Step-by-Step Solution:

Verification / Alternative check:

VR = 650 / 13 = 50. If efficiency = MA/VR = 20/50 = 0.4 (40%), the numbers are self-consistent for a real machine (optional check).

Why Other Options Are Wrong:

15, 18, 26, and 25 do not match Load/Effort; 20 is the only correct ratio.

Common Pitfalls:

Mistaking VR for MA; mixing kg (mass) and kgf (force); ignoring that MA is a pure ratio.

Final Answer:

20