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Q: As shown in the fig. a small block of mass 'm' is placed on a bigger block of mass 'M'. Friction coefficient between the blocks is mu and the surface between bigger block and surface is frictionless. Find the maximum force which can be applied on smaller box in the horizontal forward direction to prevent the relative motion between the blocks.


Q: As shown in the fig. a small block of mass 'm' is placed on a bigger block of mass 'M'. Friction coefficient between the blocks is mu and the surface between bigger block and surface is frictionless. Find the maximum force which can be applied on smaller box in the horizontal forward direction to prevent the relative motion between the blocks.

Grade:12

2 Answers

Pratham Ashish
17 Points
12 years ago

hi,

if both boxes move together their acceleration,

   a = F / (m+M)

 for big box , it will move only due to friction force

 it moves with max accel. when the friction will be max.

    a M =  µ. mg  / M                                                        ( max friction =   µ. mg  )

 if both boxes move together ,

 a =  a M

 F / (m+M) = µ. mg  / M     

 F =  (m+M) µ. mg  / M     

 it will be the max force for both the boxes move together

 

Ramesh V
70 Points
12 years ago

forces on block M and m are

on balancing forces we get,

N2 =mg

N = N2 +Mg

F-f=ma

f=Ma

so from above

F = (m+M)a

putting above gives : umg = Ma

so   F=u(M+M)mg / M

where u is coeff. of friction btn teo blocks

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