6. Block A (left side) has a mass \"m\" and block B (center) has a mass of Tefci

Question

6. Block A (left side) has a mass "m" and block B (center) has a mass of Tefcient of kinetic friction between block B and the surface is"" Block B is accelerating to the right at"a" Given [m, H, al, Determine: a. The mass of block C b. The tension in each rope 7. Block "m," sits on a horizontal frictionless surface. Block "m2 is hanging below the pulleys as shown. All of the pulleys are massless and frictionless. Given [m, m.], Determine: a. The tension in each rope. m, b. The acceleration of each block. nI

Explanation / Answer

Problem 6:

as all blocks are connected by strings, so from string constrain accelartion of all the blocks will be same.

acceleration of block A = blcok B = block C = a

Using Fnet = ma ( in vertical direction on block A)

T1 - mAg = mA*a

T1 = m(a + g) .....................Ans Tension in left side rop (between b and A)

now on B,

N - mBg = 0

N = mBg = 3mg

friction f = uk. N = 3 uk mg

in horizontal,

T2 - T1 - f = mBa

T2 - m(a + g) - 3ukmg = 3ma

T2 = 3ma + ma + mg + 3ukmg

T2 = m (4a +g +3ukg) ......... Ans tension in right side string

Now in block C,

mCg - T2 =mC*a

T2 = mC ( g - a)

m (4a +g +3ukg) = mC (g-a)

mC = m [ (4a +g +3ukg) / (g-a) ] .........Ans (MAss of block C)

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