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A block of mass 5kg is put on the top of...

A block of mass 5kg is put on the top of a 6kg block placed on a frictionless table. A horizontal force of 15N must be applied to stop the top block to cause it to slip on the bottom block as shown in Fig.. Find the maximum horizontal force F that must be applied to the lower block so that the blocks will move together and the resulting acceleration of the blocks.

Text Solution

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The force of friction between the top and bottom block is
`f_(s)=mu_(s)m_(1)g`
`mu_(s)=(f_(s))/(m_(1)g)=(15)/(5xx9.8)=0.306`
Let the acceleration of the system be a. When `a le mu_(s) g` the top block will not slip. The corresponding acceleration `a=mu_(s)g=0.306xx9.8=3ms^(-2)`.
Maximum force that can be applied on the bottom block `=F=(m_(1)+m_(2))a`
`=(5+6)xx3=33N`
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