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Inclined Plane
--------------
Friction
--------
Vertical force = mg
Normal force, FN = mgcosθ
Parallel to slope force = mgsinθ
frictional force = Ff = μFN where μ is the coefficient of friction.
mass will slide if mgsinθ > Ff with acceleration = gsinθ - Ff/m
Ex: A box slides downwards at a constant velocity on an inclined surface that has a coefficient
of friction μ = .58 The angle of the incline, in degrees, is calculated as follows:
Ff = μmgcosθ
mgsinθ ~ μmgcosθ
μ = tanθ
θ = 30 degrees
With frictionless pulley:
m2g - T = m2a2
T - m1gsinθ - μkm1gcosθ = m1a1
For simplicity consider μk = 0
Now a1 = a2 = a. Therefore, if we substitute for T we get:
T = m2g - m2a
Therefore,
m2g - m2a - m1gsinθ = m1a
Thus,
a = (m2g - m1gsinθ)/(m1 + m2)
Note: if θ = 0 the equation reduces to:
a = m2g/(m1 + m2)
Which the same as the table case.
T
m1 --->-------
------------ O
//////////| |
^ T
|
m1
F1 = T = m1a1 - μkm1g = m1a1
F2 = m2g - T = m2a2
so,
m2a2 = m2g - T
= m2g - m1a1
For simplicity assume μk = 0
Now a1 = a2 = a. Therefore,
a = m2g/(m1 + m2)
and
T = m1a = m1m2g/(m1 + m2)