Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions 185

friction is //sin a + I sin a, and the equilibrium

condition will be written in the form

1 1+ sin 2 a

sin a +sin a ) =Fir/

sin

cos a sin a cos a sin a

Ffr N = N

1+ sine a

—

sins a+ cos 2 a

1

= N

2 tan a+ cot a

On the other hand, the friction cannot exceed

the sliding friction RN, and hence

1

11 > 2 tan a+ cot a '

This inequality must be fulfilled at all values of

the angle a. Consequently, in order to find the

0 t‘

Ffr

Fig. 181

minimum coefficient of friction !Amin, we must find
the maximum of the function (2x 2 + 1/x 2 )-1, where

x 2 = tan a. The identity 2x 2 + 1/x 2 = (V^2 x —

110 -j- 2 VI implies that the maximum value of

Ni cos a= Ff r /

whence