Block B on a horizontal tabletop is attached by very light horizontal strings to
ID: 1322671 • Letter: B
Question
Block B on a horizontal tabletop is attached by very light horizontal strings to two hanging blocks, A and C. The pulleys are ideal, and the coefficient of kinetic friction between block B and the tabletop is 0.100. The masses of the three blocks are mA=12.0 kg, mB=7.0 kg, and mC=10.0 kg. Find the magnitude and direction of the acceleration of block B after the system is gently released and has begun to move. I already seperated each body and found an equation for the sum of the forces for each. They are: A-> T(1)=117.6 * 12a B-> T(sub2) - T(1) + 6.86= -7a and C-> T(2)=10a + 98 So my question is, how do I find the magnitude using these and the direction of block B? If the T1 (Tension 1, A pulling B) = 112.32 and T2 (Tension 2, B attached to C) = 102.4? I also calculated a=0.44, I believe thats correct. Block B is in between Block A and C, with A on the left, B center, and C on the right. Can someone help..?!
Explanation / Answer
For the sum of forces on block A...
mg - T1 = ma
(12)(9.8) - T1 = 12(a)
T1 = 117.6 - 12a
The sum of forces on block C
T2 - mg = ma
T2 = 98 + 10a
The sum of forces on block B
T1 - T2 - umg = ma
Sub in for T1 and T2
117.6 - 12a - 98 - 10a - .1(7)(9.8) = 7a
12.74 = 29a
a = .439 m/s2 in the direction of block A