(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.$1$. Th

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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 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 $f_{SS} = \mu_{SS} \times R = 0.57 \times 100 = 57 \ N$.$3$. The friction force on the bottom surface (between steel and brass) is $f_{SB} = \mu_{SB} \times R = 0.44 \times 100 = 44 \ N$.$4$. The total force $F$ required to move the plate is the sum of these two friction forces:$F = f_{SS} + f_{SB} = 57 \ N + 44 \ N = 101 \ N$.

Class 11 Physics Laws of Motion Kinetic Friction and Motion on Rough Horizontal Surface

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$A$ movable steel plate is placed between fixed steel and brass plates and the stack of plates is subjected to a weight of $100 \ N$ as shown in the figure. The coefficient of kinetic friction for steel on steel is $0.57$ and for steel on brass is $0.44$. Assuming that the entire weight comes onto the stack and that the weight of the plates is negligible in comparison to the applied weight, the force required to move the middle plate (in $N$) is

TS EAMCET 2018 All TS EAMCET

A
$13$
B
$101$
C
$440$
D
$570$

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Class 11

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Chapter

Laws of Motion

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Kinetic Friction and Motion on Rough Horizontal Surface

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