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A block A (5kg) rests over another block...


A block A (5kg) rests over another block B (3kg) placed over a smooth horizontal surface. There is friction between A and B. A horizontal force `F_1` gradually increasing from zero to a maximum is spplied to A so that the blocks move together without relative motion. Instead of this wanother horizontal force `F_2`, gradually increasing from zero to a maximum is spplied to B so that the blocks move together without relative motion. Then

A

`F_1(max)==f_2(max)`

B

`F_1(max)gtF_2(max)`

C

`F_1(max)ltF_2(max)`

D

`F_1(max)`:`F_2(max)=5:3`

Text Solution

Verified by Experts

The correct Answer is:
B, D
Case I:
Since, no relative motion:
`a=(f_1-F_f)/(5)=(F_f)/(3)impliesF_(1(max))=(8)/(3)F_f`
Case II: `a=(F_f)/(5)=(F_2-F_f)/(3)impliesF_(2(max))=(8)/(5)F_f`
Clearly: `F_(1(max))gtF_(2(max))` and `(F_(1(max))/(F_(2(max))=(5)/(3)`
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