Question

machine of mechanical advantage 6 is used for a load of 120 Kgf. Find the required effort. *​

Answer 1

Answer:

Given

Mechanical advantage = 6

load = 120 kgf

Mechanical Advantage = Load/Effort

Effort =load/Mechanical advantage

=120kgf/6 =20kgf

Ans=20kgf

Answer 2

Given:

Mechanical advantage = 6

Load = 120 N

To find:

Effort

Explanation:

• Mechanical advantage is simple terms is the ratio of energy produced by the system and the force applied on the system.

• Although mechanical advantage is 100% efficient for example pushing a cart, the effort given on the load is efficiently transferred to the load and the cart moves forward.

• Mathematically, mechanical advantage is

                  $Mechanical Advantage (M.A.)=$  $\frac{P_{o} }{P_{i} }$  

            where;  $P_{o}$ = Power output /Load

                          $P_{i}$ = Power input / Effort

Solution:

According to the question, we have

Mechanical advantage $(M.A.) = 6$ ; and

Load ($P_{o}$) = $120 N$

Substituting the given values in the equation, we get

Effort = $\frac{120}{6}$

         $=20N$

Final answer:

Hence, the required effort for the load of $120 N$is $20 N$.