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shubhamsrg
 3 years ago
coefficient of friction between the 2 blocks is U and the surface below the bigger block is smooth .
Find the min and max force that can be applied in order to keep the smaller blocks at rest with respect to the bigger block.
shubhamsrg
 3 years ago
coefficient of friction between the 2 blocks is U and the surface below the bigger block is smooth . Find the min and max force that can be applied in order to keep the smaller blocks at rest with respect to the bigger block.

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shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1356948181967:dw

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1356949501640:dw is that right ?

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0sorry.. this is right! missed the pseudo on second small block..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This problem was already solved on OS. Will try to find the reference.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1eqns of motion would be Umg + m(F/M) = T and T + Um(F/M) = mg => Umg + m(F/M) + Um(F/M) = mg the value of F that am getting is not right ! ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356961990631:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry that case was when it tends to move to the right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, dw:1356962249332:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Fr1= umg Fr2 =umF/(M+2m)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0From WFr2F'Fr1=0 we get F=g(M+2m)(1+u)(1u) From F'mgFr2Fr1=0 we get F =g(M+2m)(1+u)/(1u)

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1i see!! thanks a lot! that helped !

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1i understand it now..
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