Mechanics 1
Friction
Friction
Friction is a force that opposes motion and occurs when
two surfaces are in contact.
Friction
It will therefore act in the opposite direction to motion.
Laws of Friction
 When two surfaces are in contact, the force of friction opposes the
motion.
 If two surfaces are in equilibrium, the force of friction is just
sufficient to prevent motion and can be found by resolving the
forces parallel to the surface.
 Friction force will not be larger than necessary to prevent motion.
 The size of friction force is limited. Once the limited friction is
reached and the acting force is great enough to overcome friction
then motion will occur. At such point equilibrium is broken. That is
when:
Limiting Friction F = µR
The maximum value of friction is calculated by using the formula:
Fmax = µ R
Reaction
Coefficient of friction
dependant upon the material of the surface and object.
The friction acting on the particle is Fmax when :
1. The particle is in ‘limiting equilibrium’ (on the point of slipping).
2. The particle is moving.
The following equation is used for friction when an object
is at rest and is not moving.
F≤µR
NOTE : Friction can be less than or equal to it’s maximum
value.
The magnitude of the frictional force can change to
ensure equilibrium.
Friction
Fmax
As the force pulling the sleigh increases, so will the
friction to ensure equilibrium.
If the force increases to a greater magnitude than the
maximum value of friction then the sleigh will move.
(limiting friction)
Friction On A Surface
Reaction
Force
Tension
Weight
Resolving Vertically, R = mg. Where g = 9.8 m/s2
Resolving Horizontally, F = T for equilibrium at rest.
If T>F there will be a resultant force, which will produce an
acceleration in the positive direction.
Force =ma = T – F according to Newton’s Second Law of
Motion. Where F= µR.
Example 1:
A horizontal rope is attached to a crate of mass 70 kg at rest on
a flat surface. The coefficient of friction between the floor and
the crate is 0.6. Find the maximum force that the rope can exert
on the crate without moving it.
Rope
Weight
Reaction
Solution:
Given:
m = 70 kg and µ = 0.6
F < µR and R = mg
Find:
F=?
R=?
R = mg
R = 70 x 9.8 = 686 N
F < µR
F < 0.6 x 686
F < 411.6 N
Friction
Rope
Weight
Example 2:
A block of mass 15 kg rests on a rough horizontal surface. Coefficient of
friction between block and the plane is 0.25. Calculate the friction force
acting on the block when a horizontal force of 50 N acts on the block.
State whether the block will move. If it does find its acceleration.
R
F
50 N
W
Solution:
Given: m = 15 kg and µ = 0.25 with Horizontal Force = 50 N
Using: R = mg
R = 15 x 9.8
F
= 147 N
Using: F = µR
= 0.25 x 147
= 36.75 N
Since F < 50N So the block moves.
Using Newton’s Second Law of Motion:
ma = 50 – 36.75
20 x a = 13.25 N
Hence, a = 0.66 m/s2
R
50 N
W
https://www.youtube.com/watch?v=dvk7p5-nDJs
Friction On A Slope
Reaction
Weight
Note : On a slope the weight can be split into 2 components
perpendicular and parallel to the slope.
The perpendicular component of the weight pushes against the
surface so it is the Reaction that indicates the push against the
surface.
Example 3:
15o
The diagram shows an object, of mass 8 kg, on a rough plane inclined at
an angle of 15° to the horizontal.
(a) Given that the object is at rest, calculate the least possible value of
the coefficient of friction. Give your answer correct to two decimal
places.
(b) Given that the coefficient of friction is 0·1, find the acceleration of
the object.
Solution:
Given: m = 8 kg and Angle θ = 15o
a)
Resolving Perpendicular forces:
R = mg cos θ
= 8 x 9.8 x cos 15o
= 75.73N
Resolving Parallel forces:
F = mg sin θ
= 8 x 9.8 x sin 15o
= 20.29N
Using, F = µR
20.29 = µ 75.73 Hence, µ = 0.27
15o
mgcos θ W
b) Given: µ = 0.1
F = µR
= 0.1 x 75.73
= 7.573N
Using F = ma = 20.29 – 7.573
8 x a = 12.717
a = 1.59 m/s2
Example 4:
A rough plane is inclined at an angle α to the horizontal where sin α = 3/5.
A body of mass 8 kg lies on the plane. The coefficient of friction between the body and
the plane is μ.
a) Find the normal reaction of the plane on the body.
b) The body is on the point of slipping down the plane. Find the value of μ.
c) Calculate the magnitude of the force acting along a line of the greatest slope that will
move the body up the plane with an acceleration of 0.7 m/s2.
Solution:
Given: sin α = 3/5 and m = 8kg
α = sin-1 (3/5) = 36.9o
a)
Resolving Perpendicular forces:
R = mg cos θ
= 8 x 9.8 x cos 36.9o
= 62.72N
b)
Resolving Parallel forces:
F = mg sin θ
= 8 x 9.8 x sin 36.9o
= 47.04N
Using, F = µR
47.04 = µ 62.72 Hence, µ = 0.75
c)
Given: a= 0.7 m/s2
F = 8 x 0.7 = T – (47.04 + 47.04)
T = 99.68N
Example 5:
A particle is held at rest on a rough plane which is inclined to the
horizontal at an angle α where tan α = 0.75. The coefficient of friction
between the particle and the plane is 0.5. The particle is released and
slide down the plane. Find,
a) The acceleration of the particle.
b) The distance it slides in the first 2 seconds.
Solution:
Given: tan α = 0.75 and µ = 0.5
=> α = tan-1 (0.75) = 36.87o
a)
Resolving Perpendicular forces:
R = mg cos θ
= mgcos 36.87o
Resolving Parallel forces:
F = mg sin θ
= mg sin 36.87o
Using, F = µR
= 0.5 x mg cos36.87o
Using, Newton’s Second Law of Motion:
ma = mg sin 36.87o - 0.5 x mg cos36.87o
Diving by m on both sides:
a = 9.8 x sin 36.87o - 0.5 x 9.8 x cos36.87o
a = 1.96 m/s2
b)
Given t = 2 seconds
Using,
S = ut + ½ at2
Since the particle slides from rest.
Hence, U = 0 m/s2
S = 0 x t + ½ x 22 x 1.96
= 3.92 m