Question

Figure shows the use of a lever. If $FA=10cm,AB=500cm$, $load=\text{5 kgf}$, the minimum effort required to lift the load is :

• A. $2500N$
• B. $3550N$
• C. $2550N$
• D. $2520N$

Answer 1

Answer: (C)

$2550N$

$FA=EffortArm=10cm$
Load Arm $=FA+AB=10+500=510cm$
Mechanical Advantage $=\frac{\text{Effort Arm}}{\text{Load Arm}}=\frac{10}{510}=\frac{1}{51}$
Load $=5kgf$
Therefore, the minimum effort required to lift the load is:

Effort $=\frac{\text{Load}}{\text{Mechanical Advantage}}=\frac{5}{\frac{1}{51}}=5\times 51=255kgf=2550N$