In the figure a painter of mass 100 kg pulls himself up with the crate of 25 kg with an acceleration. If the painter exerts an effective force of 450 N on the floor of the crate, find (a) the acceleration of the painter and
(b) the tension in the string taking `g=10 m//s^(2)`

Explanation

Suppose the man pulls the rope with constant force `ved(F)` downwards then equation of motion for point B will be
`F+m_(B)g-T=m_(B)a`
i.e., `F=T[as m_(B)=0]` ...(i)
Whiel for painter
`R+T-mg=ma`
`i.e., R+T=m(g+a)` ….(ii)
And for the system
`2T-(m+M)g=(m+M)a`
i.e., `2T=(m+M)(g+a)` ....(iii)
Here `m=100 kg M=25 kg, R=A=450 N`
and `g=10 m//s^(2)`
So, from eqns. (ii) and (iii) we have
`450+T=100(10+a) and 2T=(100+25)(10+a)`
i.e., `T=550 + 100a and 2T=125(10+a)`
which of solution yields
`a=2 m//s^(2) and T=F=750N [ "as from eqns "(i) F=T]`
Note: (i) The equation of motion for the crate will be
`T-A-Mg=Ma`
i.e., `T-R=M(g+a)[as A=R]` ....(iv)
and if we add eqns (ii) and (iv) we get
`2T=(m+M)(g+a)`
i.e., the equation of motion of system (iii) which cosists of painter and crate
(ii) Free diagram is:

(iii) From Eqns (ii) and (iii) `R-(m-M)(g+a)//2` so the problem is physically possible only if
`R ge0 " i.e., "m geM`