21YEAH!!!!!!!!!! k k...... Finally I got it!!!! [132]
THANKS A LOT subhash n Kalyan!!! [1]
11from your diag(considering m and M for boy and cart)
120-Fr=ma(eqn for boy)
Fr=ma( eqn for cart)
(a is same for both to avoid sliding)
Normal reaction between the two bodies N=mg
so Fr≤μN
take equality for minimum
applying values you can get the ans
1the reaction of friction is responsible only for the acceleration of the cart.
And the cart weighs 10 kg[1]
1If f is the friction
(120-f)N= 30a
f=10a
a=f/10
120-f=3f
4f=120
f=30
f=μN
=> μ=f/30g
=0.1
1If f is the friction
(120-f)N= 30a
f=10a
a=f/10
120-f=3f
4f=120
f=30
f=μN
=> μ=f/30g
=0.1
21oh k...
so
N = 40g, a =120/(M+m) = 3, M = 30, m =10;
but then F = 90 = μ(400)
then μ≠0.1 [2]
21Ok...
So 120N = 40a
a = 3
Friction = μN
N = 30g
we gotta equate Friction to ?? ma kya??
