Question

An effort of 4 N is applied to lift a load of 20 N. If the effort arm is 5m, calculate the following: (i) load arm (ii) input work (iii) output work​

Answer 1

Answer:

An effort of 4 N is applied to lift a load of 20 N. If the effort arm is 5m.

(i) load arm         =   1 m

(ii) input work     =   20 J

(iii) output work​  =   20 J

Explanation:

Given,

Load = 20 N           Effort = 4 N

Effort arm = 5m

Mechanical advantage (M.A) can be obtained by calculating the ratio of load and effort.

That is,  

M.A =   $\frac{Load}{Effort}$

        =   $\frac{20}{4}$ = 5

$\text { Effort } \times \text { effort arm }=\text { load } \times \text { load arm }$

$\text { Load } \mathrm{arm}=\frac{\text { Effort arm } \times \text { Effort }}{\text { Load }}$

Substituting the values,

 $\text { Load } \mathrm{arm}=\frac{\text { 5 } \times \text { 4 }}{\text { 20 }}$ = 1 m

We have  the formula for velocity ratio as,

Velocity ratio = $\frac{Effort arm}{Loadarm}$   =  $\frac{5}{1} = 5$

Efficiency can be written as,

Efficiency , $\eta= \frac{\text { M.A. }}{\text { V.R. }}\times100$  %  = $\frac{5}{5} \times100= 100$ %

Also, Efficiency =  $\frac{workoutpu}{work inpu}\times 100$ %

In this case, efficiency is 100% which means this is an ideal machine.

so, Output work = input work.

Output work =   Load $\times$ Load distance

                     =   20  $\times$  1 = 20 J

Iput work      =  $\frac{outpu work}{efficiency}$ = $\frac{20}{1}$ = 20J