11.0.0. FORCE:
Force is an agent which change the states of an object. It is an agent which either causes a stationary object to move or stop a moving object. Force is a vector quantity.
UNIT OF FORCE:
The s.I unit of force is newton. Other unit of force is kgm/s square.
TYPES OF FORCE:
Force is classified into two:
• Contact Force:
This is a force that act directly on the object. This type of force is always in contact with the it is acting upon.
EXAMPLE OF CONTACT FORCE:
(i) Tension force (ii) force of push and force of pull (iii) friction force.
• FIELD FORCE:
This is a type of force that act on an object from a distance. This type of force is not in contact with the object it is acting upon neither does it act directly on the object.
EXAMPLE OF FIELD FORCE
(i) Magnetic force (ii) electric force or force from a charge object (iii) gravitational force.
FRICTION FORCE:
Friction force is the type of force which exist between two surfaces in contact and tries to oppose their relative motion. The force start to exist when two surfaces want to move over one another.
DIRECTION OF FRICTION FORCE:
Friction force act in a perpendicular direction to the two surfaces in contact. It come to existence when one surface tries to move over the other surface.
TYPES OF FRICTION FORCE:
• Static or Limiting Frictional Force:
This is the maximum amount of friction force that must be overcomed before an object can just start to move over another object.
• Kinetic or dynamic friction force :
This is the friction force that must be overcomed so that a moving object can continue to move with constant speed over another object.
Note that coefficient of static friction is always greater than coefficient of dynamic friction.
Coefficient of Static Friction Force > Coefficient of Dynamic Friction Force
LAWS OF SOLID FRICTION
The laws of solid friction states that :
• Friction force exist between two surfaces in contact
• Friction force opposes the relative motion of the two surfaces in contact
• Friction force is proportional to normal reaction
• Friction force does not depend on the area of the surfaces in contact
• Friction force act in a direction that is opposite to that of motion
• Friction force depends on the nature of the two surfaces in contact
RELATIONSHIP BETWEEN FRICTION FORCE, NORMAL REACTION AND COEFFICIENT OF FRICTION:
From the definition,
Friction force is directly proportional to normal reaction
Friction force & normal reaction
APPLICATION / ADVANTAGES OF FRICTION FORCE:
• Friction force makes movement possible
• Friction make car brake to function
• Friction is used in grinding stone
• Friction enable screws and nail to remain in place after screwed or nailed
• Friction is used in belt and pulley system
DISADVANTAGES OF FRICTION FORCE:
• Friction force reduce efficiency of machine
• Friction force generate heat
• Friction causes wear and tear of machine parts
• Friction reduces motion
• Friction causes wastage of capital
METHOD OF REDUCING FRICTION:
• Streamlining of object:
Streamlining of objects means to shape an object with a pointed end. Streamlining helps to reduces friction force between an object and air or water molecules.
• Lubrication:
Lubrication of surfaces in contact helps to separate the two contact surfaces and therefore eliminate or reduces friction force between the two surfaces. The lubricant form a thin film between the surfaces and therefore eliminate friction force between the surfaces.
• Using ball bearing or rollers:
Ball bearing between surfaces separate then surfaces and therefore eliminate friction force between the surfaces.
• Publishing of surfaces:
Publishing of surfaces helps to make the surfaces smooth which in turn eliminate friction force between the surfaces.
• Introducing air between the surfaces:
Introducing air between the surfaces help to separate the surfaces and therefore reduces Friction force between surfaces
EXPERIMENTAL DETERMINATION OF COEFFICIENT OF FRICTION USING BLOCK AND TABLE:
Diagram or Setup:
DETERMINATION OF COEFFICIENT OF FRICTION USING INCLINE PLANE: The coefficient of friction can also be determined using incline plain.
Apparatus:
Incline plain, block of mass,
DIAGRAM / SET UP
➡ 
PROCEDURES:
Place the block on the incline plane as shown in the figure above. Gradually increase the angle of inclination of the plane until the object is just about to slide down the plane.
Read and record the angle of inclination of the inclined plane as θ and the mass of the block as M.
Let W be the weight of the block. W = mg.
EXPLANATION:
From the diagram, mg is the weight of the block. The weight of the object act vertically downward as indicated by the vertical line and has two components:
Fperp ( force perpendicular to the plane ) is mg x Cos θ which produced the normal reaction .
Fpara is the mg x Sin θ which make the block to slide along the incline plane.
Fg is the weight of the block which act vertically downward.
Note:
the formulae are obtained from the concept of SOH CAH TOA.
VECTOR TRIANGLE:
The force that move the object parallel to the plane or along the plane
Sin θ = force that act along the plane (Fm) ÷ weight of object
force that act along the plane (Fm) = weight x Sin θ
Fm = W x sin θ = mg x Sin θ ( W = mg )
We will also find the weight that act in a perpendicular direction to the plane surface:
Cos θ = Normal reaction R ÷ weight of object W
Normal reaction R = Cos θ x Weight of object = Cos θ x W
R = mg x Cos θ
From the law of friction force,
Force F is directly proportional to Normal reaction R
F = µ & R , if we make µ the subject of the equation, then we have
µ = F / R
At this point, we will substitute for F and R .
therefore, µ = F / R , µ = mg x Sin θ / mg x Cos θ
µ = tan θ
µ is used to represent coefficient of friction force.
APPLICATIONS:
Example 1.
A 10kg mass is placed on an in line plane in lined at an angle of 30° .
Calculate: I) The force parallel to the plane. II) the force normal to the plane
II) The coefficient of friction
Solution:
Diagram of incline plane:
Angle of inclination θ = 30°, mass = 10kg,
weight = mass x acceleration due to gravity. That is, w = mg
From the vector diagram above
• Sin θ = Force / weight of object , Force = Sin θ x W ,
Force = Sinθ x mg, Force = Sin 30° x 10 x 9.8, Force = 0.5 x 98
There Force parallel to the plane = 46Newton
Also,
• Cosθ = R / W,
R = Cosθ x W, R = Cos θ x mg
* You must take Note that R is called Normal reaction. It is the force that is perpendicular to the plane surface. Ok! if we substitute for R and W,
Therefore, R = Cos 30° x 10 x 9.8 , R = 0.8660 x 98, R = 84.87Newton
Therefore, force perpendicular to the plane surface= 84.87 N
• Coefficient of friction µ = F / R = 46 / 84.87 = 0.54
* You must also note that “the force that act along / parallel to the plane” is the force that will move the object along the plane while the force that is perpendicular to the plane surface is the normal reaction R. Ok!
EXERCISES:
• A block of mass 1.2kg is placed on an incline plane at 60° to the horizontal. Calculate: a) the force that will make the block to slide down the plane. b) the coefficient of friction. c) the normal reaction, d) the friction force.
• An object of mass 10kg is placed on an incline plane of angle of inclination 30°. Calculate : a) the reaction between the two surfaces. b) the coefficient of static friction.
• A mass 25kg is placed on a horizontal table with coefficient of friction 0.5. If the mass is static, determine: a) the weight of the object b ) the limiting friction force.
• A body just rest at equilibrium on a slop of incline plane at angle ® where Sin® = 4 /15. Find the value of Π
• If the coefficient of friction is 0.5 and reaction is 70N. Calculate: a) limiting friction force that prevent the object from sliding ( g = 10m/s^2)
• Two bodies P and Q, 12kg and 18kg are connected by a light inextensible string as shown below. P is placed on a rough surface of coefficient of friction 0.50. If the acceleration of P is 10m/s^2, calculate: a) he normal reaction on P. b) the friction force between P and the surface. C) the tension in the string. D). The force which caused P and Q to accelerate. (SSCE Nov/December)
• 30N force applied parallel to the surface of a table is just enough to make the block of mass 4kg move. Calculate: a) coefficient of friction between the block and the table. ( g = 10m/s^2) ( SSCE June 1992)
• 8kg mass of block on a table accelerates at 0.1m/s^2 when a force is applied. Calculate the coefficient of friction between the table and the block. (SSCE 1998)
Force is an agent which change the states of an object. It is an agent which either causes a stationary object to move or stop a moving object. Force is a vector quantity.
UNIT OF FORCE:
The s.I unit of force is newton. Other unit of force is kgm/s square.
TYPES OF FORCE:
Force is classified into two:
• Contact Force:
This is a force that act directly on the object. This type of force is always in contact with the it is acting upon.
EXAMPLE OF CONTACT FORCE:
(i) Tension force (ii) force of push and force of pull (iii) friction force.
• FIELD FORCE:
This is a type of force that act on an object from a distance. This type of force is not in contact with the object it is acting upon neither does it act directly on the object.
EXAMPLE OF FIELD FORCE
(i) Magnetic force (ii) electric force or force from a charge object (iii) gravitational force.
FRICTION FORCE:
Friction force is the type of force which exist between two surfaces in contact and tries to oppose their relative motion. The force start to exist when two surfaces want to move over one another.
DIRECTION OF FRICTION FORCE:
Friction force act in a perpendicular direction to the two surfaces in contact. It come to existence when one surface tries to move over the other surface.
TYPES OF FRICTION FORCE:
• Static or Limiting Frictional Force:
This is the maximum amount of friction force that must be overcomed before an object can just start to move over another object.
• Kinetic or dynamic friction force :
This is the friction force that must be overcomed so that a moving object can continue to move with constant speed over another object.
Note that coefficient of static friction is always greater than coefficient of dynamic friction.
Coefficient of Static Friction Force > Coefficient of Dynamic Friction Force
LAWS OF SOLID FRICTION
The laws of solid friction states that :
• Friction force exist between two surfaces in contact
• Friction force opposes the relative motion of the two surfaces in contact
• Friction force is proportional to normal reaction
• Friction force does not depend on the area of the surfaces in contact
• Friction force act in a direction that is opposite to that of motion
• Friction force depends on the nature of the two surfaces in contact
RELATIONSHIP BETWEEN FRICTION FORCE, NORMAL REACTION AND COEFFICIENT OF FRICTION:
From the definition,
Friction force is directly proportional to normal reaction
Friction force & normal reaction
Friction force = µ x normal reaction,
µ = friction force / normal reactionAPPLICATION / ADVANTAGES OF FRICTION FORCE:
• Friction force makes movement possible
• Friction make car brake to function
• Friction is used in grinding stone
• Friction enable screws and nail to remain in place after screwed or nailed
• Friction is used in belt and pulley system
DISADVANTAGES OF FRICTION FORCE:
• Friction force reduce efficiency of machine
• Friction force generate heat
• Friction causes wear and tear of machine parts
• Friction reduces motion
• Friction causes wastage of capital
METHOD OF REDUCING FRICTION:
• Streamlining of object:
Streamlining of objects means to shape an object with a pointed end. Streamlining helps to reduces friction force between an object and air or water molecules.
• Lubrication:
Lubrication of surfaces in contact helps to separate the two contact surfaces and therefore eliminate or reduces friction force between the two surfaces. The lubricant form a thin film between the surfaces and therefore eliminate friction force between the surfaces.
• Using ball bearing or rollers:
Ball bearing between surfaces separate then surfaces and therefore eliminate friction force between the surfaces.
• Publishing of surfaces:
Publishing of surfaces helps to make the surfaces smooth which in turn eliminate friction force between the surfaces.
• Introducing air between the surfaces:
Introducing air between the surfaces help to separate the surfaces and therefore reduces Friction force between surfaces
EXPERIMENTAL DETERMINATION OF COEFFICIENT OF FRICTION USING BLOCK AND TABLE:
Apparatus:
Smooth table surface, pulley, block of wood, weight, rope and spring balance
DETERMINATION OF COEFFICIENT OF FRICTION USING INCLINE PLANE: The coefficient of friction can also be determined using incline plain.
Apparatus:
Incline plain, block of mass,
DIAGRAM / SET UP
➡ 
PROCEDURES:
Place the block on the incline plane as shown in the figure above. Gradually increase the angle of inclination of the plane until the object is just about to slide down the plane.
Read and record the angle of inclination of the inclined plane as θ and the mass of the block as M.
Let W be the weight of the block. W = mg.
EXPLANATION:
From the diagram, mg is the weight of the block. The weight of the object act vertically downward as indicated by the vertical line and has two components:
- The weight (mg) of the block produce the force that causes the block to move downward alone the plane or parallel to the plane. The force is mg x Sin θ
- The weight of the block produced a vertical force which act perpendicularly to the plane surface. The force is mg x Cosθ
- f represent the friction force that tries to oppose the force the cause the block to move downward along the plane.
- N is the normal reaction produced by plane surface on the block due to
- mg x Cosθ
Fperp ( force perpendicular to the plane ) is mg x Cos θ which produced the normal reaction .
Fpara is the mg x Sin θ which make the block to slide along the incline plane.
Fg is the weight of the block which act vertically downward.
Note:
the formulae are obtained from the concept of SOH CAH TOA.
VECTOR TRIANGLE:
The force that move the object parallel to the plane or along the plane
Sin θ = force that act along the plane (Fm) ÷ weight of object
force that act along the plane (Fm) = weight x Sin θ
Fm = W x sin θ = mg x Sin θ ( W = mg )
We will also find the weight that act in a perpendicular direction to the plane surface:
Cos θ = Normal reaction R ÷ weight of object W
Normal reaction R = Cos θ x Weight of object = Cos θ x W
R = mg x Cos θ
From the law of friction force,
Force F is directly proportional to Normal reaction R
F = µ & R , if we make µ the subject of the equation, then we have
µ = F / R
At this point, we will substitute for F and R .
therefore, µ = F / R , µ = mg x Sin θ / mg x Cos θ
µ = tan θ
µ is used to represent coefficient of friction force.
APPLICATIONS:
Example 1.
A 10kg mass is placed on an in line plane in lined at an angle of 30° .
Calculate: I) The force parallel to the plane. II) the force normal to the plane
II) The coefficient of friction
Solution:
Diagram of incline plane:
Angle of inclination θ = 30°, mass = 10kg,
weight = mass x acceleration due to gravity. That is, w = mg
From the vector diagram above
• Sin θ = Force / weight of object , Force = Sin θ x W ,
Force = Sinθ x mg, Force = Sin 30° x 10 x 9.8, Force = 0.5 x 98
There Force parallel to the plane = 46Newton
Also,
• Cosθ = R / W,
R = Cosθ x W, R = Cos θ x mg
* You must take Note that R is called Normal reaction. It is the force that is perpendicular to the plane surface. Ok! if we substitute for R and W,
Therefore, R = Cos 30° x 10 x 9.8 , R = 0.8660 x 98, R = 84.87Newton
Therefore, force perpendicular to the plane surface= 84.87 N
• Coefficient of friction µ = F / R = 46 / 84.87 = 0.54
* You must also note that “the force that act along / parallel to the plane” is the force that will move the object along the plane while the force that is perpendicular to the plane surface is the normal reaction R. Ok!
EXERCISES:
• A block of mass 1.2kg is placed on an incline plane at 60° to the horizontal. Calculate: a) the force that will make the block to slide down the plane. b) the coefficient of friction. c) the normal reaction, d) the friction force.
• An object of mass 10kg is placed on an incline plane of angle of inclination 30°. Calculate : a) the reaction between the two surfaces. b) the coefficient of static friction.
• A mass 25kg is placed on a horizontal table with coefficient of friction 0.5. If the mass is static, determine: a) the weight of the object b ) the limiting friction force.
• A body just rest at equilibrium on a slop of incline plane at angle ® where Sin® = 4 /15. Find the value of Π
• If the coefficient of friction is 0.5 and reaction is 70N. Calculate: a) limiting friction force that prevent the object from sliding ( g = 10m/s^2)
• Two bodies P and Q, 12kg and 18kg are connected by a light inextensible string as shown below. P is placed on a rough surface of coefficient of friction 0.50. If the acceleration of P is 10m/s^2, calculate: a) he normal reaction on P. b) the friction force between P and the surface. C) the tension in the string. D). The force which caused P and Q to accelerate. (SSCE Nov/December)
• 30N force applied parallel to the surface of a table is just enough to make the block of mass 4kg move. Calculate: a) coefficient of friction between the block and the table. ( g = 10m/s^2) ( SSCE June 1992)
• 8kg mass of block on a table accelerates at 0.1m/s^2 when a force is applied. Calculate the coefficient of friction between the table and the block. (SSCE 1998)
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