More than a Pulley Problem...

The max force that the peg can take is given by 360 N/ sin 30= 720 N.

Anand with weight 600 N thus has 120 N to use for accelerating upwards. F= ma

120=60*a

a = 2 m/s^2

Using equation of motion,

s= ut +1/2 at^2

81= t^2

t = 9 seconds.

For Anand to be held stationery, a Tension force of 600N vertically upwards must be sustained. Since the rope over the pully is one rope, we treat the tension in it equally. so the tension pulling on the Peg would be 600. If we get the 2 components by resolution, the vertical force is:

$\\ 600\sin { 30 } \quad =\quad 300N\\ so\quad the\quad force\quad Aanand\quad can\quad apply\quad can\quad only\quad affect\quad \\ the\quad peg\quad by\quad a\quad maximum\quad of\quad 60\quad N.\quad But\quad remember\quad \\ that\quad any\quad force\quad he\quad applies\quad is\quad going\quad to\quad be\quad multiplied\quad \\ by\quad \sin { 30 } \quad =\quad \frac { 1 }{ 2 } .\quad This\quad translates\quad to:\\ X\sin { 30 } =\quad 60.\quad \\ X\quad =\quad 120\quad N\\ We\quad then\quad use\quad F\quad =\quad ma\quad to\quad get\\ 120\quad =\quad 60a\\ a\quad =\quad 2\quad m{ s }^{ 2 }\\ Now\quad using\quad Kinematics\\ u\quad =\quad 0,\quad a\quad =\quad 2,\quad s\quad =\quad 81,\quad t\quad =\quad z\\ s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } a{ t }^{ 2 }\\ Substituting\quad in\quad Values:\\ 81\quad =\quad t^{ 2 }\\ t\quad =\quad 9\quad seconds\\$

The total force on Anand is Fu:

Fu = Ft - Fw;

by the force Fu Anand accellerate upward,

Fw is the weight force of Anand, Ft the total tension on the rope;

Fv = 360 N is the maximum vertical vincular reaction of the peg,

Ft = Fu + Fw = ma + mg = m(a + g); (1)

for the equilibrium on the peg:

Fv/sinθ > Ft; Ftsinθ < Fv; (with θ = 30°) (2)

the height h is:

h = (1/2)at^2; so: a = 2h/(t^2); (3)

combining (1) and (3) in (2) we obtain:

m(2h/(t^2) + g)sinθ < Fv; which leads to:

t > √(2h/(Fv/(msinθ) - g));

• Hypotesis: g ~ 10 m/s; Fv = 360 N; m = 60 Kg; θ = 30°; h = 81 m;

t > √(2 x 81/(360/(60 x 1/2) - 10)) = √(2 x 81/(720/60 - 10));

t > √(2 x 81/(12 - 10)) = √(2 x 81/2) = 9 = tmin (ANSWER)

Hi am anand :D

Answer is sqrt(2 81/360 60) = 9 seconds

The maximum tension on the rope should be 720N [360/cos60] Anand weight is 600N, so he can pull hisself with 120N up. F=m . a, 120 = 60 . a, a=2 Y= vt +1/2 at^2, 81=1/2.2t^2, t= 9sec.