Answer:
Answered
Explanation:
v= 1 m/s
A= 1 m^2
m= 100 kg
y= 1 mm
μ = ?
ζ= viscosity of SAE 20 crankcase oil of 15° C= 0.3075 N sec/m^2
forces acting on the block are
F_s ← ↓ →F_f
mg
N= mg
F_s= shear force = ζAv/y F_f= friction force = μN
now in x- direction F_s= F_f
ζAv/y = μN
0.3075×1×1×1/1×10^{-3} = μ×100
⇒μ=0.313 (coefficient of sliding friction for the block)
Now, as the velocity is increased shear force also increases and due to this frictional force also increases.
Now, to compensate this frictional force friction coefficient must increase
as v∝μ
Answer:
The coefficient of sliding friction is 3.13
Solution:
As per the question:
Mass of the block, m = 100 kg
Velocity of the block, v = 1.0 m/s
Thickness of the oil layer, t = 0.10 cm = [tex]0.10\times 10^{- 3}\ m[/tex]
Temperature, T = [tex]15^{\circ}C[/tex]
For SAE 20crankcase oil, [tex]\mu = 0.3075\ Ns/m^{2}[/tex]
Area, A = [tex]1.0\times 1.0 = 1.0\ m^{2}[/tex]
Now,
To calculate the sliding friction coefficient:
[tex]f = \mu_{s}N = \mu_{s} mg[/tex]
Oil friction, [tex]f = \tau A[/tex]
where
[tex]\tau[/tex] = Torque
A = Area
Also,
[tex]\tau = \mu \frac{du}{dy}[/tex]
[tex]\tau = \mu \frac{v}{t}[/tex]
Therefore,
[tex]f = \mu_{s} \frac{v}{t} A[/tex]
[tex]\mu\frac{v}{t} A = \mu_{s} mg[/tex]
[tex]\mu_{s} = {0.3075\times \frac{1.0}{0.1\times 10^{- 3}}\times 1}{100\times 9.8} = 3.13[/tex]
