Question

minimum force required to move a body of an inclined plane of inclination 30° found to be twice the minimum force required to prevent the sliding the plane the coefficient of friction VP ​

Answer 1

HELLO DEAR,

minimum force required to move the body up on a inclined place is mgsin30 + umgcos30.

so, minimum force required to prevent from sliding is mgsin30 - umgcos30.

GIVEN:- twice the minimum force required to prevent the sliding the plane the coefficient of friction VP

therefore,

mgsin30 + umgcos30 = 2(mgsin30 - umgcos30)

=> 3umgcos30 = mgsin30

=> tan30 = 3u

=> u = (tan30)/3

=> u = (1/√3)/3

=> u = 1/3√3

therefore, the cofficient of friction is (1/3√3).

I HOPE IT'S HELP YOU DEAR,

THANKS

Answer 2

Answer:

minimum force required to move the body up on a inclined place is mgsin30 + umgcos30.

so, minimum force required to prevent from sliding is mgsin30 - umgcos30.

GIVEN:- twice the minimum force required to prevent the sliding the plane the coefficient of friction VP

therefore,

mgsin30 + umgcos30 = 2(mgsin30 - umgcos30)

=> 3umgcos30 = mgsin30

=> tan30 = 3u

=> u = (tan30)/3

=> u = (1/√3)/3

=> u = 1/3√3

therefore, the cofficient of friction is (1/3√3).