A box of mass 50 kg is placed on an inclined plane. When the angle of the plane is increased to 30 degree, the box begins to slide downwards. Calculate the coefficient of static friction between the plane and the box.
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body starts to slide at 30 degree
hence ,
mg.sin30 =uN
where u is coefficient of static friction and N is normal reaction between point of contact .
N=mgcos30
now,
mgsin30 =umgcos30
u=tan30 =1/root3
hence coefficient of static friction=1/root3
hence ,
mg.sin30 =uN
where u is coefficient of static friction and N is normal reaction between point of contact .
N=mgcos30
now,
mgsin30 =umgcos30
u=tan30 =1/root3
hence coefficient of static friction=1/root3
SaurabhEinstein:
can u make a free fall diagram of it
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