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`        if the minimum force required to drag a heavy block is n times its own magnitude when the force is applied in an inclination of coefficient of friction with the horizon.find coefficient of friction.`
8 years ago

pratham ashish
9 Points
```										hi,

in the situation u mention in ur qn ,

f up   nmg sin(a)               tan(a)=mu (given in qn)

f down   mg

f towards rt  nmg cos(a)
f towards lft   mu*mg(1-nsin(a))   {total dwnward f*mu}

if this is the min. then,  f  lft = f rt =>  mu*mg(1-nsin(a))= nmg cos(a)

=> mu = ncos(a)/{1-nsin(a)}

put sin(a)= mu/sqr root{1-mu^2}

&  cos(a)= 1/sqr root{1-mu^2
then cross multiply u wud get

u^2{1+u^2}=n^2*{(1+u^2)}^2
=>   u^2 =n^2* {1+u^2}
=>    1/n^2 = 1/u^2 +1
=>   u = n/sqr root{1-n^2}.

```
8 years ago
Ramesh V
70 Points
```										I have solved the answer in one of your previous posts..plzz follow through this link:
```
8 years ago
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