1.

The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. If the angle of inclination of the plane is 60^(@), then the coefficient of friction is

Answer»

`(1)/(3)`
`(1)/(SQRT(2))`
`(1)/(sqrt(3))`
`(1)/(2)`

Solution :According to the QUESTION, angle of inclination `theta = 60^(@)`.

Now , force of friction, f = `mu`N = `mu` mg cos `theta`
and net retarding force, `(F_(1)) ` = mg sin `theta` + f
`therefore` Net accelerating force down the inclined plane is
F = mg sin `theta - mu mg cos theta"" ` .... (i)
`therefore` EXTERNAL force NEEDED (up the inclined plane ) to maintain sliding motion is (net retarding force )
`F_(1) = mg sin theta + mu mg cos theta "" `....(ii)
it is given as, `F_(1) = 2F""`.... (iii)
From Eqs. (i) , (ii) and (iiI) we get
mg sin `theta` + `mu mg cos theta` = 2 (mg sin `theta - mu mg cos theta` )
or `3mu mg cos theta = mg sin theta `
or `""(sin theta)/(cos theta) = 3mu`
or `"" tan theta = 3 mu "" (because theta = 60^(@))`
or `"" tna 60^(@) = 3mu`
or `"" sqrt(3) = 3mu`
or `"" mu = (1)/(sqrt(3))`
When inclination of plane is `60^(@)` then the coefficient of friction , `mu = (1)/(sqrt(3))`


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