A slide on friction

2.  Friction is a tangential force acting parallel to two surfaces in contact and
moving in respect to one another.
 Force does not cause movement but it is a RESULT of movement.
 There are no perfectly frictionless surfaces in engineering application.
There are two types of friction : 1. Kinetic
2. Static
 Static Friction- friction that resists a stationary body to move
 Kinetic Friction- friction that resists movement of a body that is already
moving

3.  The ratio of frictional force to the normal force which acts perpendicular to
the frictional force.
 𝜇 =
𝐹
𝑁
; F is the frictional force, N is the normal force perpendicular to F, µ is
the coefficient.

4.  This is the angle between the resultant force of F and the normal reaction.
R is the resultant; Fn is the normal reaction;
Ft is the frictional force, is the angle of friction.
The angle between the surface and the horizontal when
The body is about to slide is known as the angle of repose(Ф)
R
Ф

5. Law1- Friction is proportional and perpendicular to the normal force.
𝐹 = 𝜇𝑁
∴ 𝐹 ∝ 𝑁
Law2- Friction is independent of area of contact, but instead is dependant on
the coefficient and the normal Force.
Law3- coefficient of static friction is slightly greater than the coefficient of
kinetic friction.
Law4- Friction depends on the nature of the surfaces involved.

6. 1. Draw Free body diagrams.
2. Show Frictional force in the correct direction.
3. Determine the number of unknowns
4. Do not assume F=µN unless told the impending motion condition is given.
5. Apply equilibrium equations and appropriate frictional equations to
determine the unknowns.

7. If the coefficient of kinetic friction between a 35-kg crate and the floor is 0.30, what
horizontal force is required to move the crate to the right at a constant speed across the
floor?
mg
Fn
Fa
Ff
(0.30)(35)(9.8)
a f f k N
a k N
N
a k
a
a
F F F F
F F
F mg
F mg
F
F



 




 102.9 N

8.  A person pushes a 30-kg shopping cart up a 10 degree incline with a force of 85 N. Calculate
the coefficient of friction if the cart is pushed at a constant speed.
cosmg 
mg
Fn
Fa
Ff
sinmg 
0.117


sin
sin cos
cos sin
sin cos
sin
cos
85 (30)(9.8)(sin10)
(30)(9.8)(cos10)
a f f k N
a k N N
a k
a k
a
k
k
F F mg F F
F F mg F mg
F mg mg
F mg mg
F mg
mg
 
  
  
  




  
  
 
 



 

9. A ladder of length 4 m, weighing 200 N is placed against a vertical wall as
shown in Fig. 5.14(a). The coefficient of friction between the wall and the ladder
is 0.2 and that between floor and the ladder is 0.3. The ladder, in addition to
its own weight, has to support a man weighing 600 N at a distance of 3 m from
A. Calculate the minimum horizontal force to be applied at A to prevent
slipping.

10. Solution: The free body diagram of the ladder is as shown in Fig. 5.14(b).
ΣMA = 0 →
200 x 2 cos 60 + 600 x 3 cos 60 – FB x 4 cos 60 – NB x 4 sin 60 = 0
Dividing throughout by 4 and rearranging the terms, we get
0.866 NB + 0.5 FB = 275
From law of friction, FB = NB = 0.2 NB
∴ 0.866 NB + 0.5 0.2 NB = 275
or NB = 284.7N
∴ FB = 56.94N
ΣFV = 0 →
NA – 200 – 600 + FB = 0
NA = 743.06 N, since FB = 56.94
∴ FA = μANA
= 0.3 743.06 = 222.9N
ΣFH = 0 →
P + FA – NB = 0
∴ P = NB – FA = 284.7 – 222.9
P = 61.8N

11. 𝑇2 = 𝑇1 𝑒 𝜇𝜗

12.  The sliding surfaces are present in most machine components (bearings,
gears , cams, etc.) and it is desirable to minimize the friction in order to
reduce energy loss and wear.
 In contrast, clutches and brakes depend on friction in order to function.
 The function of a clutch is to permit smooth, gradual connection and
disconnection of two elements having a common axis of rotation.
 A brake acts similarly except that one of the elements is fixed.

13. Journal bearings provide lateral support to rotating
shafts. Thrust bearings provide axial support
• Frictional resistance of fully lubricated bearings
depends on clearances, speed and lubricant viscosity.
Partially lubricated axles and bearings can be assumed
to be in direct contact along a straight line.
• Forces acting on bearing are weight W of wheels and
shaft, couple M to maintain motion, and reaction R of
the bearing.
• Reaction is vertical and equal in magnitude to W.
• Reaction line of action does not pass through shaft
center O; R is located to the right of O, resulting in a
moment that is balanced by M.