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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.
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
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
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 --- Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
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
---
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
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