Just before slipping, we say that the block is still at rest, so the net force on the x-axis is: EFx=mg sintheta -f s Evaluating, we get: mg sintheta But the frictional force is expressed as the product of coefficient and the normal force so. Hs N = mg For an inclined plane. The normal force is n=mg Then, Ps=mg sintheta cos(e) so: Ps- 1 tan(e)

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A block of mass m is placed on a rough surface inclined relative to the horizontal. The incline angle is increased until the block start to move. Show that
you can obtain the coefficient of static friction ug by measuring the critical angle 0 where slipping occurs.
mg pino
to
....
my
18.
Transcribed Image Text:Problem A block of mass m is placed on a rough surface inclined relative to the horizontal. The incline angle is increased until the block start to move. Show that you can obtain the coefficient of static friction ug by measuring the critical angle 0 where slipping occurs. mg pino to .... my 18.
Solution
Just before slipping, we say that the block is still at rest, so the net force on the x-axis is:
EFx=mg sintheta
-f s
Evaluating, we get:
= mg sintheta
But the frictional force is expressed as the product of coefficient and the normal force so.
Hs N
= mg
For an inclined plane. The normal force is
n=mg
Then,
Ps=mg sintheta
cos(e)
so:
Ps= 1
tan(e)
Transcribed Image Text:Solution Just before slipping, we say that the block is still at rest, so the net force on the x-axis is: EFx=mg sintheta -f s Evaluating, we get: = mg sintheta But the frictional force is expressed as the product of coefficient and the normal force so. Hs N = mg For an inclined plane. The normal force is n=mg Then, Ps=mg sintheta cos(e) so: Ps= 1 tan(e)
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