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pranay -askiitians expert Grade: 12
        

6.     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 –


7jul10-img1.jpg


plz explain it. answer is 10 N

6 years ago

Answers : (2)

vikas askiitian expert
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...


 

6 years ago
pranay -askiitians expert
44 Points
										

but the answer was given as 10 n i got the same answer

6 years ago
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