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Friction

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The coefficient of friction is $\\mu_s=\\var{mu_s}$ between all surfaces of contact. Knowing that $M_1=\\var{M1}$kg and $M_2=\\var{M2}$kg, determine the smallest force $\\mathbf{P}$ required to start the block $M_2$ moving if:

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(a)

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Free body: $\\var{M1}$-kg block

\n

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$W_1=M_1 g=\\var{m1}\\times 9.81=\\var{m1*9.81}\\,\\text{N}$

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$+\\!\\!\\uparrow\\sum F_y=0:\\quad N_1=W_1$

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$F_1=\\mu_s N_1=\\var{mu_s}\\times\\var{m1*9.81}=\\var{mu_s*m1*9.81}\\,\\text{N}$

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$\\overset{+}{\\rightarrow}\\sum F_x=0:\\quad T-F_1=0,\\quad T=F_1=\\var{mu_s*m1*9.81}\\,\\text{N}$

\n

\n

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Free body: $\\var{M2}$-kg block

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$W_2=M_2g=\\var{M2}\\times 9.81=\\var{m2*9.81}\\,\\text{N}$

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$+\\!\\!\\uparrow\\sum F_y=0: \\quad N_2-N_1-W_2=0,\\quad N_2=W_1+W_2=\\var{(m1+m2)*9.81}\\,\\text{N}$

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$F_2=\\mu_s N_2=\\var{mu_s}\\times\\var{(m1+m2)*9.81}=\\var{mu_s*(m1+m2)*9.81}\\,\\text{N}$

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$\\overset{+}{\\rightarrow}\\sum F_x=0: -P+F_1+F_2=0$

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\\[P=F_1+F_2=\\var{P1}\\,\\text{N}\\quad\\blacktriangleleft\\]

\n

\n

(b)

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Blocks move together (same as in Question 4)

\n

\\[P=\\var{P2}\\,\\text{N}\\quad\\blacktriangleleft\\]

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cable AB is attached as shown, 

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cable AB is removed.

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