33Tricky question :P
Wait
A block of mass 1kg held by a sting is on an inclined plane of inclination 37 degrees. The coeff. of friction (μ) = 0.8. Find the tension in the string.
-
UP 0 DOWN 0 0 12
12 Answers
33If it is allowed to slide when string was loose and then equillibrium is achieved then T=mg*sin(37)-μmgcos(37)
33Its a bit tricky..
before the contact with inclined string was tight and inclined at the angle of incline and then kept on plane then T=mgsin(37)
1if the block is in motion,then there shud be a frictional force acting on it....in this case is it equal to μmgcosθ??
33Second case is only a guess or like it :P
Not physically possible...
See imagine some how U were able to switch of the gravity in perpendicular to plane..
Contd..
33and keep only mg sin(37) on so T= mg sin(37) and keep it on plane(inclined) and then turn on normal component of gravity.. then it is case 2..
it is only to show that some more description in question was needed and i think u were confused about it only as this is otherwise a straight forward question...
But if this comes in exam case 1 will be answer most probably..
33So 1st case...
See there friction is sufficient to balance mg sin(37) so static friction and T=0..
I didn't looked at the data.. I told you the method...
See if friction frm case 1 t must be coming out to negative which can't be possible so static friction..
@Nishant Bhaiya
U were talking of such cases naa where we must take inequality but T<0 gave a hint :P