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```
Block A of mass 2 kg is placed over a Block B of mass 8 kg. The combination is placed on a rough horizontal surface. If g=10 ms^-2, coefficient of friction between B and floor= 0.5, coefficient of friction between A and B = 0.4 and a horizontal force of 10N is applied on 8kg block, Then the force of friction between A and B is?

```
6 years ago

```							f1+f2 =10N
f1=friction b/w floor and B
f2 = friction b/w B and A
m1=8Kg; m2=2kg
(mew)1=0.5
(mew)2=0.4
f1=R(mew)1
f1=(m1+m2)g(mew)1
f1=(2+8)×10×0.5
f1=50N
f1+f2=10N
f2=-40N
-ve due to opposition
```
6 years ago
```							1st equation will be f1+f2=10Nf1 force of friction between a and b blockf2 force of friction between b and surfacef1=8Nf2=50NTherefore the force applied 10N is less than the frictional force of 50N. So the block will not move and the friction is limiting friction. And before limiting there is  static friction which  is  a self adjusting force.Static friction is -f1
```
3 years ago
```							From above  firstly check maximum friction at ground  =0.5×(8+2)g =0.5×100 =50N Since 50N>10N the applied so there is no friction between A and B
```
2 months ago
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