In fig. the man and the platform together weight 950N. The pulley can be treated as frictionless. Determine how hard the man has to pull on the rope to lift himself upward above the ground with constant velocity. If the weight of man is 550 N, what is the normal reaction between them?

Answer

20 kg wt; 40 kg wt; 80 kg wt.

Explanation

Let N be the reaction between man and platform. `T_(1)` is the tension in string passing over pulley 1, `T_(2)` is the tension in string passing over pulley 2. M is the mass of man, m is the mass of platform.
The free-body diagram of man platform and pulley 2 are shown in fig.


Action exerted by man of platfrom =reaction exerted by platform on man =N
For equilibrium of man: `Mg=T_(2)+N` ..(i)
For equilibrium of platform: `T_(1)+T_(2)=M+Mg` ...(ii)
For equilibrium of pulley 2, `T_(1)=2T_(2)` ...(iii)
Adding Eqs. (iii) we get
`T_(1)+T_(2)+(T_(2)+N)=N+mg+Mg`
`T_(1) +2T_(2) =(M +m)g`
Using Eq. (iii), we get `2 T_(2) + 2T_(2) = (M+m)g`
`implies T_(2)=((M+m))/(4)g`
`:. F=T_(2)=((M+m))/(4)g=((60+20)/(4)) g N = 20 kg wt`.
Now From eq (i)
`N=Mg-T_(2)=Mg-F=Mg-((M+m))/(4)g1`
`=((3M-m))/(4)g` ...(v)
`=((3 xx 60 -20)/(4)) g = 40 kg wt`
Form Eq. (v), `mg=3 Mg-4N`
Clearly mg is maximum when N is minimum.
Minimum `N=0`
so, `(mg)_(max)=3 Mg=3 xx 60 g = 180 kg wt`.