# 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?

anonymous
20 Points
10 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
Umang Jain
11 Points
7 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
ankit singh
4 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