Question

A resistance force is 225 N and the AMA is 1.74. What is the effort force?

Answer 1

Answer:

Effort force is given as 129.3 N

Explanation:

As we know that AMA is the ratio of out force and input force

so here we will have

$AMA = \frac{F_{out}}{F_{in}}$

here we know that

$F_{out} = 225 N$

$F_{in} = 1.74$

now we have

$1.74 = \frac{225}{F_{in}}$

so we have

$F_{in} = 129.3 N$