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

Options

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

Answer

B

Explanation

For given figure,
`mu_(SS)=` coefficient of friction between steel and steel = 0.57
and `mu_(SB) =` coeifficient of friction between steel and brass = 0.44.

where, `f_(SS)` =friction between steel plates and `f_(SB)`= friction between steel and brass plates.
`rArr` For kinetic-friction, the applied force (F) is equal or greater than friction force for moving the steel plate, so `f_(SS) + f_(SB)le F`
Therefore, `f_(SS)=mu _(SS)xx R = 0.57xx100=57N`
`f_(SB)=mu_(SB)xxR=0.44xx100=44N`
`rArr` 57 + 44 = N
`rArr` 101 newton = N