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```        .     The two blocks, m = 10kg and m = 50 kg are free to move as shown. The coefficient of static friction between the blocks Is 0.5 and there is no friction between M and the ground. A minimum horizontal force F is applied to hold m against M that is equal to                              please explain eah and every step

```
7 years ago

## Answers : (1)

510 Points
```										F = (M+M)a
a=F/m+M ............1
bw m and M there is normal reaction let it be N,
then  at m
F-N=ma
N=F-ma  or
=F(1 - m/(m+M) )
N=F( M/m+M)................2
if both blocks are at rest wrt each other then it can be said that friction is sufficient to balance the weight of  block of mass m..
(friction force) f =mg
f =u(N)                    (u is cofficient of friction and N is normal reaction)
uN=mg
u(F)(M/m+M) =mg
F=mg(m+M)/uM
on solving
F=240N
this is the minimum force required for which block remmains in equilibrium...
if force if more then friction is more and then also block remains in equilibrium,so 240N is the value is minimum possible value of force...
```
7 years ago
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