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Introductory Physics Homework Help
Two Blocks and a Pulley with Friction
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[QUOTE=". Arctic., post: 4399561, member: 467310"] [h2]Homework Statement [/h2] Blocks A and B are connected by a cable that passes over support C. Friction between the blocks and the inclined surfaces may be neglected. Knowing that motion of block B up the incline is impending when m[SUB]B[/SUB] = 4.5 kg, determine: (a) the coefficient of static friction between the rope and the support (b) the largest value of m[SUB]B[/SUB] for which equilibrium is maintained [h2]Homework Equations[/h2] T[SUB]2[/SUB] = T[SUB]1[/SUB]e^(μ[SUB]s[/SUB]β) (1) β is in angle in radians W = mg [h2]The Attempt at a Solution[/h2] My problem is that for part (b), my answer is wrong. I would like to know where I went wrong in my calculations. Thanks for the help in advance. I found the forces working on Block A ƩF[SUB]x[/SUB] = 0 → T[SUB]A[/SUB] - W[SUB]A[/SUB]sin(16) = 0 T[SUB]A[/SUB] = 24.336 N Then, Block B ƩF[SUB]x[/SUB] = 0 → W[SUB]B[/SUB]sin(16) - T[SUB]B[/SUB] = 0 T[SUB]B[/SUB] = 12.168 N Then, I looked at support C where I found the angle to be 32°, and I used (1) to find the static friction. β = (32)(∏/180) ≈ 0.559 T[SUB]A[/SUB] = T[SUB]B[/SUB]e^(μ[SUB]s[/SUB]β) 24.336 N = (12.168 N)e^(μ[SUB]s[/SUB]β) e^(μ[SUB]s[/SUB]β) = (24.336 N)/(12.168 N) μ[SUB]s[/SUB]β = ln [(24.336 N)/(12.168 N)] μ[SUB]s[/SUB] = (1/0.559)*ln [(24.336 N)/(12.168 N)] μ[SUB]s[/SUB] ≈ 1.240 Then, since I know T[SUB]B[/SUB] has to be greater than itself to upset the equilibrium, I did T[SUB]B[/SUB] = T[SUB]A[/SUB]e^(μ[SUB]s[/SUB]β) T[SUB]B[/SUB] = (24.336 N)e^[(1.240)(0.559)] T[SUB]B[/SUB] ≈ 48.673 N Then, to get the mass I divided the T[SUB]B[/SUB] by g = 9.81 m[SUB]B[/SUB] = (48.673)/(9.81) ≈ 4.962 kg That is wrong. [/QUOTE]
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Two Blocks and a Pulley with Friction
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