Question

flat block is pulled along a horizontal flat surface by a horizontal rope perpendicular to one of the sides. The block measures 1.0 m × 1.0 m, has a mass of 100 kg and a constant velocity of 1.0 m/s, and is separated from the flat surface by a 0.10-cmthick oil layer of 15 °C SAE 20 crankcase oil. Find the coefficient of sliding friction for the block. Does this coefficient of friction change with the velocity?

Answer 1

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 2

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 = $0.10\times 10^{- 3}\ m$

Temperature, T = $15^{\circ}C$

For SAE 20crankcase oil, $\mu = 0.3075\ Ns/m^{2}$

Area, A = $1.0\times 1.0 = 1.0\ m^{2}$

Now,

To calculate the sliding friction coefficient:

$f = \mu_{s}N = \mu_{s} mg$

Oil friction, $f = \tau A$

where

$\tau$ = Torque

A = Area

Also,

$\tau = \mu \frac{du}{dy}$

$\tau = \mu \frac{v}{t}$

Therefore,

$f = \mu_{s} \frac{v}{t} A$

$\mu\frac{v}{t} A = \mu_{s} mg$

$\mu_{s} = {0.3075\times \frac{1.0}{0.1\times 10^{- 3}}\times 1}{100\times 9.8} = 3.13$

• Sliding friction depends directly on the block's velocity and hence changes with the velocity of the block.