1It wud have been so if both were stationary... but the monkey is moving upwards
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).
-
UP 0 DOWN 0 0 10
10 Answers
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....
1
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 ?