Sum
moments about the center of the pulley.
Σ Mcenter = 0 TA = TB =
T
Sum
forces in x-direction for block A.
Σ Fx = mA aA T
- F =
mA aA (1)
Sum
forces in y-direction for block A.
Σ Fy =
0 WA - N =
0 so N
= WA (2)
Sum
forces in y-direction for block B.
Σ Fy = mB aB WB - T =
mB aB (3)
Now
the maximum friction force F =
μ N = μ WA
(4)
From
kinematics (for inextensional cord) aA =
aB =
a
(5)
Put (4) into (1) and add to (3) gives WB – μ WA
= (mA
+ mB) a
So a
= ( WB – μ WA
) / (mA + mB)
= ( 16.1 – 6.44 ) / ( 2 + 1 )
> 0 So
sliding occurs.
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