Two cubes of masses `m_1` and `m_2` be on two frictionless slopes of block A which rests on a horizontal table. The cubes are connected by a string which passes over a pulley as shown in figure. To what horizontal acceleration f should the whole system (that is blocks and cubes) be subjected so that the cubes do not slide down the planes. What is the tension of the string in this situation?

Answer

`T=(m_(1)m_(2)g sin(alpha-beta))/(m_(1)cos alpha+m_(2)cos beta)`

Explanation

If the block has an acceleration towards right,the blocks would have some acceleration towards left. Resoving horizontally and vertically, we have
`R_(2)=m_(2)g cos beta +m_(2)f sin beta .....(1)`
`m_(2)f=Tcos beta +R_(2) sin beta ......(2)`
and
`R_(1)=m_(1)g cos alpha+m_(1)fsin alpha .....(3) m_(1)f=R_(1) sin alpha alpha - T cos alpha ......(4)`
From equaiton `(2)` and `(1)` , we get
`m_(2)f=T cos beta +[m_(2)g cos beta +m_(2)fsinbeta]sin beta`
`=T cos beta +m_(2)g cos beta sin beta +m_(2)f sin^(2)beta`
`:. T cos beta =m_(2)f-m_(2)fsin^(2)beta-m_(2)g cos beta sin beta `
`or T cos beta m_(2) cos^(2) beta-m_(2)beta cos beta sin beta`
`or T=m_(2)fcos beta -m_(2)g sin beta .....(5)`
From equations `(3)` and `(4)` , we have
`m_(1)f=(m_(1)gcos alpha+m_(1)f sin alpha)sin alpha-T cos alpha`
`=m_(1)g cos alpha sin alpha+m_(1)f sin^(2)alpha-T cos alpha `
`:. T cos alpha=m_(1)g cos alpha sin alpha +m_(1)f(sin^(2)alpha-1)`
`or T =m_(1)g sin alpha-m_(1)f cos alpha ....(6)`
Equating `(5)` and `(6)`, we get
`m_(2)f cos alpha -m_(2)g sin beta=m_(1)g sin alpha-m_(1)f cos alpha `
`:. f=g^(g)[((m_(1)sinalpha+m_(2)sinbeta))/((m_(2)cos beta+m_(1)cos alpha))] ....(7)`
Substituting the value of `f` in `eqn. (6)`, we get
`T=m_(1)g sin alpha - m_(1)g`
`{((m_(1)sin alpha+m_(2)sinbeta))/((m_(2)cos beta+m_(1)cos alpha))}cos alpha`
Simplifying , we get
`T=(m_(1)m_(2)gsin(alpha-beta))/(m_(1)cosalpha+m_(2)cos beta)Ans.`