written 8.7 years ago by | • modified 4.9 years ago |
Determine the velocity of each block when system is started from rest & block B gets displacement by 2m. Take µk=0.2 between block A & horizontal surface.
written 8.7 years ago by | • modified 4.9 years ago |
Determine the velocity of each block when system is started from rest & block B gets displacement by 2m. Take µk=0.2 between block A & horizontal surface.
written 8.7 years ago by |
Consider length of string w.r.t. block A as XA &
Length of string w.r.t block B as XB
So, applying C.S.L.M. we get
2XA+XB=C
But as B moves down, XB increases & XA decreases
So, rewriting eqn
−2XA+XB=C______(i)
Differentiating w.r.t. t
−2VA+VB=0............2VA=VB_____(ii)
Differentiating once more w.r.t. t
2aA=aB____(iii)
Applying NSL to block A
∑Fx=max & ∑Fy=0T−0.2(N)=10aA N=10gT−0.2(10g)=10aA_____(iv)
Applying NSL to block A
∑Fy=may5g−T=maB_____(v)
Solving eqns (iii)(iv)&(v),we get
T=34.335NaA=1.4715m/s2aB=2.934m/s2
Now, after displacement of 2m, let velocity of B be VB
(VB)2=0+2aB(2)....VB=3.43m/s
Putting in eqn (ii), VA=1.715m/s
Velocity if A is 1.715m/s & Velocity of B is 3.43m/s