Loads of 2kg and 10 kg are hanging from an ideal thread over a smooth pulley. The system is in equilibrium...when a monkey weighing 8kg starts climbing up from the given position at time t=0 with an acceleration 2m/s2 relative to the string. Find the time in which the monkey reaches the top of the pulley (h=2m).
1let rel accln be a0, accln ok 10 kg be 'a' upwards & accln of 2 kg dwnwds :
m0 is mass of monkey
T-m0g=m0(a0-a)
T-m2g=m2a
Subtract both......a= -2/5
therefore, accln of monkey wrt ground=a0-a= 2+2/3=12/5
time = root (2*dist)/root (a0-a) = root(5/3) ... dist moved is given 2m.
11I sould not get either of u.....I mean what are ur reference frames?
Bcoz if u take a ref frame fixed to the string, then the mass attacthed to it will be at rest!
3won't the accln of the strng b zero i mean both the bloks will nt acclr8 at all ...there is 10 kg blok on the left side and 10 kg (8(monkey) +2) on the right side also....so the relative velocity of the monkey w.r.t strng will be his velocity w.r.t grnd also....
1It wud have been so if both were stationary... but the monkey is moving upwards
1Bhaiyon aur Beheno , Devi aur Sajjanon here.....remember dat equilibrium is distorted due to the monkey . Tension upto monkey and tension below monkey r different. Ab tension chhoro aur monkey pe dhyaan do.....sum ho jayga.. :-)
1what are the length of the string on both sides of the pulley ?