Basic Concept: When two bodies are in contact, each one is exerting a force on the other. Therefore, friction can be defined as the force which tends to oppose the relative sliding motion of two surfaces in contact. Frictional force is the opposing force between two forces in contact.
The direction of friction is opposite to the direction in which the motion will occur. The frictional force for any two surfaces in contact has a value given by”
F = UR
Where R is the normal reaction between the bodies and obtained as; R = m x g
U = Coefficient of friction and the value depends only on the nature of the surfaces in contact.
Examples
1: A mass of 4kg rests on a rough horizontal table, with U = 0.4. Find the least force sufficient to move the mass: (take g = 10ms-2)
Solution
Sufficient force; F = UR.
R = m x g = 4 x 10 = 40
F = 0.4 x 40
F = 16N
2. If a force of 10N is just sufficient to move a mass of 2kg resting on a rough horizontal table, find the coefficient of friction (g = 10ms-2)
Solution
F = 10 N , m = 2kg.
F = UR
10 = U x 2 x 10
10 = 20U
u = 10/20
U = 0.5
EVALUATION
A body of mass 8kg rests on a horizontal surface. If the coefficient of friction between the body and the horizontal surface is 0.65, calculate the minimum horizontal force enough to just move the body.
FORCES ACTING ON A BODY PLACED ON A ROUGH INCLINED PLANE
Rough Inclined Plane: There are two basic forces acting on an object placed on an inclined plane; the applied forces P and the frictional force F P= Mg Sin Ө
F = Mg SinӨ
R = Mg Cos Ө
Recal F = UR
:. F = UMgCos Ө where F is the limiting
Friction:
The least force required in P to make the body move up the inclined plane r is P = UR + Mg Sin Ө
:. P = F + MgSinӨ
P = UR + MgSinӨ
The least force required to make the body slide down the plane is thus
P + f = mgSin Ө
P = mg Sin Ө- Ur.
Where m = mass and Ө= angle of friction.
The value of the coefficient of fricton from F = UR and
MgSinӨ = UMgCos Ө is defined by
U = mg sin Ө / Mg cos Ө
:. U = tan Ө
Examples
- A block is placed on an inclined plane at an angle of 30o to the horizontal and just remains at rest. Find the coefficient of friction.
Solution
Ө = 30o, U = tan Ө
U = tan 30o
U = 1/√3 or 0.5774.
- A body of mass 4.5kg rests on a smooth plane inclined at an angle of 47o to the horizontal. Calculate the magnitude of the force P parallel to the plane just enough to prevent the body from sliding down the plane
(g= 9.8ms-2).
Solution
P = mg Sin Ө
P = 4.5 x 9.8 x sin 47o = 44.1 x 0.7314.
P = 32.25N.
EVALUATION
A particle of mass 25kg slides down a rough plane inclined at angle 30o to the horizontal. If the coefficient of friction is 0.2, find, in ms-2 the acceleration of the particle correct to 3 significant figures (take g = 10ms-2)
GENERAL EVALUATION
A uniform ladder of length 6m and mass 30kg rests with one end against a rough vertical wall and the other end on a rough horizontal ground. The coefficient of friction at each point of contact is 0.3. If the ladder is on the point of slipping, calculate the (i) normal reaction of the ground (ii) frictional force at the wall (iii) angle of inclination of the ladder to the ground. (Take g = 10ms-2)
READING ASIGNMENT
Read Friction page 191 -195
WEEKEND ASSIGNMENT
1. A body is in limiting equilibrium on a plane inclined at an angle to the horizontal. if Cos = 0.8, calculate the coefficient of friction.
2. A big stone is of mass 13kg. A boy whose weight is 59N sits on the stone. Find the minimum horizontal force P required to move the stone on the ground. If the coefficient of friction is 0.25 (g = 9.8ms-2).
3. A mass of 8kg tests on a rough table with U = 0.6. Find the least force which will make the mass move (g = 10ms-2).
4. A mass of 10kg slides down a rough plane inclined at Ө to the horizontal where sin Ө = 0.6. if U = 0.3, find the acceleration of the box.
5. A body can just rest in equilibrium on a slope inclined at Ө to the horizontal where sinӨ = 5/13, find U
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