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(B) To move the middle steel plate,the applied force $F$ must overcome the kinetic friction forces acting on both its top and bottom surfaces.$1$. The

Vedclass · Accepted Answer

$101$

Vedclass · Answer

(B) To move the middle steel plate,the applied force $F$ must overcome the kinetic friction forces acting on both its top and bottom surfaces.$1$. The normal force $R$ acting on the surfaces is equal to the applied weight,$R = 100 \ N$.$2$. The friction force on the top surface (between steel and steel) is $f_{SS} = \mu_{SS} 	imes R = 0.57 	imes 100 = 57 \ N$.$3$. The friction force on the bottom surface (between steel and brass) is $f_{SB} = \mu_{SB} 	imes R = 0.44 	imes 100 = 44 \ N$.$4$. The total force $F$ required to move the plate is the sum of these two friction forces:$F = f_{SS} + f_{SB} = 57 \ N + 44 \ N = 101 \ N$.

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