1.

In the given figure, a long platform of mass m is placed on frictionless surface. Two blocks of masses 4 m and m (where m=10 kg) are placed on the platform. For both blocks, the coefficient of static friction with platform equal to 0.16 and the coefficient of kinetic friction equal to 0.10 the blocks are connected by a light ideal string through a light pulley (mounted at a movable massless stand). Which is acted upon by an unknown horizontal force F. if the acceleration of the platform is 2m//s^(2). Find the value of unknown force F and acceleration of blocks 4 m and m on a_(1) and a_(2) respectively.

Answer»

`{:(F(N),,a_(1)(m//s^(2)),,a_(2)(m//s^(2))),(180N,,2m//s^(2),,8m//s^(2)):}`
`{:(F(N),,a_(1)(m//s^(2)),,a_(2)(m//s^(2))),(180N,,8m//s^(2),,2m//s^(2)):}`
`{:(F(N),,a_(1)(m//s^(2)),,a_(2)(m//s^(2))),(90N,,2m//s^(2),,8m//s^(2)):}`
`{:(F(N),,a_(1)(m//s^(2)),,a_(2)(m//s^(2))),(90N,,8m//s^(2),,2m//s^(2)):}`

SOLUTION :
Both block 4 m and m has tendency of motion towards rightwards so friction `(F_(1)` and `F_(2))` will act on both blocks LEFTWARD, hence on the platform rightward.So
`a=(F_(1)+F_(2))/(m)=2m//s`
`F_(1)+F_(2)=20N`
`F_(1)le64N_(1)" "F_(2)le16N`
Case I: Suppose blok of mass 4 m has a RELATIVE motion with respect to platform So, `F_(1k)=40N`
`F_(2)=-20N` (not possible)
Case II: Suppose block of mass m has a relative motion with respect to platform, so, `F_(2k)=10NimpliesF_(1s)=10N` (possible)
`(F)/(2)-F_(1s)=4mxx2implies(F)/(2)=1+4xx10xx2=90`
`impliesF=180Nimplies(F)/(2)-F_(2k)=maimplies90-10=10aimpliesa=8m//s^(2)`


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