block B on a horizontal tabletop is attached by very light horizontal strings to

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.

Explanation / Answer

for A

ma*g - Ta = ma*a

Ta = ma*g - ma*a.........(1)

for C

Tc - mc*g = mc*a

Tc = mc*a + mc*g................(2)

for B

Ta - Tc - uk*mb*g = mb*a

ma*g - ma*a - mc*a - mc*g - uk*mb*g = mb*a

a = (ma*g -mc*g -uk*mb*g)/(ma+mb+mc)

a = ((12*9.8)-(10*9.8)-(0.1*7*9.8))/(12+7+10)

a = 0.44 m/s^2

Ta > Tb (T1 > T2)

the resultant is towards left i.e along T1

B moves towards left

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