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Inclined Plane

Friction

Vertical force = mg
Normal force, F_{N} = mgcosθ
Parallel to slope force = mgsinθ
frictional force = F_{f} = μF_{N} where μ is the coefficient of friction.
mass will slide if mgsinθ > F_{f} with acceleration = gsinθ  F_{f}/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:
F_{f} = μmgcosθ
mgsinθ ~ μmgcosθ
μ = tanθ
θ = 30 degrees
With frictionless pulley:
m_{2}g  T = m_{2}a_{2}
T  m_{1}gsinθ  μ_{k}m_{1}gcosθ = m_{1}a_{1}
For simplicity consider μ_{k} = 0
Now a_{1} = a_{2} = a. Therefore, if we substitute for T we get:
T = m_{2}g  m_{2}a
Therefore,
m_{2}g  m_{2}a  m_{1}gsinθ = m_{1}a
Thus,
a = (m_{2}g  m_{1}gsinθ)/(m_{1} + m_{2})
Note: if θ = 0 the equation reduces to:
a = m_{2}g/(m_{1} + m_{2})
Which the same as the table case.
T
m_{1} >
 O
////////// 
^ T

m_{1}
F_{1} = T = m_{1}a_{1}  μ_{k}m_{1}g = m_{1}a_{1}
F_{2} = m_{2}g  T = m_{2}a_{2}
so,
m_{2}a_{2} = m_{2}g  T
= m_{2}g  m_{1}a_{1}
For simplicity assume μ_{k} = 0
Now a_{1} = a_{2} = a. Therefore,
a = m_{2}g/(m_{1} + m_{2})
and
T = m_{1}a = m_{1}m_{2}g/(m_{1} + m_{2})