D.59 [See also D.33.] Let\'s reanalyse the problem with three blocks, each of ma
ID: 1579624 • Letter: D
Question
D.59 [See also D.33.] Let's reanalyse the problem with three blocks, each of mass M. Two of them ride on a horizontal plane, while the third hangs from a pulley as shown in the figure with D.33. Assume that the system is in motion (to the right and down). Determine the (i) acceleration of the system and (ii) tension in each of the ropes in the following cases. (a) Kinetic friction, with coefficient H acts on the leftmost block only (b) Kinetic friction acts on both blocks with precisely the same coeffcient, H (c) Kinetic friction acts on both blocks with differing coefficients: Hi andExplanation / Answer
(A) on hanging block:
M g - T2 = M a
on mass on table,
T2 - uK M g = (M + M)a
adding both.
M g - uk M g = 3 M a
a = g (1 - uk)/3 ....Ans(i)
(ii) T2 = M g - Mg/3 + uk Mg /3
T2 = M g (2 + uk)/3 ....Ans
T1 - uk M g = M a
T1 = uk Mg + Mg/2 - uk Mg / 3
T1 = M g (2 uk + 1)/3
(B) on hanging block:
M g - T2 = M a
on mass on table,
T2 - uK 2M g = (M + M)a
adding both.
M g - uk 2M g = 3 M a
a = g (1 - 2uk)/3 ....Ans(i)
(ii) T2 = M g - Mg/3 + 2uk Mg /3
T2 = 2 M g (1 + uk)/3 ....Ans
T1 - uk M g = M a
T1 = uk Mg + Mg/2 - 2uk Mg / 3
T1 = M g ( uk + 1)/3
(C) on hanging block:
M g - T2 = M a
on mass on table,
T2 - uk1 M g - uk2 M g = (M + M)a
adding both.
M g - uk1 M g - uk2 M g = 3 M a
a = g (1 - uk1 - uk2 )/3 ....Ans(i)
(ii) T2 = M g - Mg/3 + uk1 Mg /3 + uk2 M g /3
T2 = M g (2M + uk1 + uk2 )/3 ....Ans
T1 - uk1 M g = M a
T1 = uk1 Mg + Mg/3 - uk1 Mg / 3 - uk2 Mg /3
T1 = M g (2 uk1 + 1 - uk2 )/3
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