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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

Grade:12

## 1 Answers

vikas askiitian expert
509 Points
11 years ago

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

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