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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`
7 years ago

510 Points
```										let the force applied on the block M of mass 50 kg is F...
then
F=(m+M)a
a=F/(m+M)................1
now upper block will experience a seudo force of magnitude ma...
due to this seudo force it will tend to move but its  magnitude should be
greater than or equal to friction force...
seudo force>=friction force
ma >=umg
a >=ug ............2
from eq 1
F >= u(m+M)g
F >=294N
so the minimum force is 294N

```
7 years ago
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