A box of uniform density with a square base of side length b and height 2b sits
ID: 1261229 • Letter: A
Question
A box of uniform density with a square base of side length b and height 2b sits on an incline. The coefficient of static friction between the box and the incline is mu s. The angle of the incline is increased until the box either tips over onto its side, or slides down the incline.
(a) For what relation between mu s and heta does the box slide? For what relation between mu s and heta does the box tip?
(b) If mu s = 0.55 does the box slide down the slope or tip? At what angle of the incline does it do this?
Please state specific..
A box of uniform density with a square base of side length b and height 2b sits on an incline. The coefficient of static friction between the box and the incline is mu s. The angle of the incline is increased until the box either tips over onto its side, or slides down the incline. (a) For what relation between mu s and theta does the box slide? For what relation between mu s and theta does the box tip? (b) If mu s = 0.55 does the box slide down the slope or tip? At what angle of the incline does it do this? Please state specific..Explanation / Answer
force due to friction = ?s * normal Force
?s = coefficient of friction
so normal force due to gravity = m * g * cos (?)
force due to friction = ?s * m * g * cos (?)
force due to gravity = m * g * sin (?)
for box to slide the force due to gravity must be greater than friction force so,
m * g * sin (?) > ?s * m * g * cos (?)
for box to slide tan (?) > ?s
for box not to tip
it should not slide that is frictional force should be equal to gravitational force
that is,
m * g * sin (?) = ?s * m * g * cos (?)
for box not to slide tan (?) = ?s
also tan (?) > (b/2)/(2b/2)
tan (?) > 0.5
or,
? = 26.57 deg
since tan (?) = ?s
so,
?s > 0.5
so for ?s = 0.55 the box will tip
It will tip if the angle is > 26.57 deg
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