LEARNING OBJECTIVESBy the end of this chapter, you should be able to:1. (a) Define: – Force. (b) State and define: the SI unit of force.2. List the types of forces.3. State the effects of force.4. (a) Define acceleration due to gravity (b) Explain why acceleration due to gravity varies from place to place (c) Describe an experiment to determine acceleration due to gravity, g.5. (a) Define: – Mass and Weight. (b) State: – The difference between mass and weight. (c) Use: – W = mg, to solve numerical problems.6. Define: – Scalar and vector quantities and give examples for each. ecolebooks.com 7. Solve numerical problems involving combination vectors (forces).

5.1  Force

Force  is either a push or pulls that acts on an object. Or

Force is that which changes a body’s state of rest or uniform motion in a straight line.

A force cannot be seen and described like an object, but the effects of force on an object can be seen. Just like we can not see wind but we can feel and see the effects in a flying flag or kite or in a tree bending.

(a)  SI Unit of force

The SI unit of force is called Newton (N) and is defined as follows:

A Newton is the force required to give a unit mass (mass of one kilogram) an acceleration of one metre per Second Squared.

1. Representation of force

Force is a vector quantity i.e. it has both magnitude and direction.

It is represented by a straight line with an arrow at the tip showing the direction in which it acts.

(c)  Effects produced by a Force

A force acts in a particular direction and may have any of the following effects on an object.

It can:  (i)  Make a stationary object to move.

(ii)  Increase the speed of a moving object.

(iii)  Decrease or slow down the speed of a moving object or bring a moving object to a rest.

(iv)  Change the direction of a moving object.

(v)  Deform (change the shape of) an object.

(d)  Types of Forces

There are various types of forces; some of the common ones include the following:

(i)  Gravitational force  – is the force which pulls a body towards the centre of

the earth.

(ii)  Frictional force  – is the force which opposes relative motion between

two surfaces
in contact.

(iii)  Centripetal force  – is a force which constrains a body to move in a

circular path or orbit. Its direction is towards the

centre of the circle.

(iv)  Magnetic force  – is a force in magnets that causes motion as a result

of attraction or repulsion.

(v)  Electrostatic force  – is a force that causes attraction or repulsion in an

electric field due to static charges.

(vi)  Elastic force – is a force in a stretched spring or rubber cord.

(vii)  Upthrust – is an upward force that acts on a body immersed in

a fluid (a liquid or a gas).

(viii)  Cohesive force  – is a force of attraction between molecules of the

same kind. E.g force of attraction between water

molecules.

(ix)  Adhesive force  – is a force of attraction between molecules of different

kinds. E.g force of attraction between water

molecules and glass molecules.

(x)  Surface tension  – is the force on a liquid surface which makes the

liquid surface to behave like a stretched elastic

membrane or skin.

5.2  (a)  Gravitational force

An object released from a height falls down to the ground. This indicates that there is a force acting on the object directed to the centre of the earth. This downward pull is called gravitational force.

Gravitational force is the force which pulls a body towards the centre of the earth.

It makes unsupported objects to fall until they reach the ground. The force of gravity pulls all objects at a particular place on the surface of the earth with the same acceleration of free fall called acceleration due to gravity.

(b)  Acceleration due to gravity, g

Acceleration due to gravity, g, is the acceleration with which the force of gravity pulls all objects on or near the surface of the earth towards its centre.

The average value of g
on the surface of the earth is 9.8 ms-2, but it varies slightly from place to place.

(c)  The variation of g

The acceleration due to gravity is not constant everywhere. It varies from place to place. For example g is 9.78 ms-2 at the equator, whereas at the poles it is 9.83 ms-2. There are two main causes of this variation.

(i)  Variation of g with altitude

Acceleration due to gravity depends on the distance from the centre of the earth. The earth is not a perfect sphere. It is oval in shape; as such places are far away from the centre of the earth while others are near to the centre of the earth. For example the equatorial radius of the earth is greater than the polar radius. That is the distance from the centre of the earth to the equator is greater than the distance from the centre of the earth to the pole. Therefore a body at the equator is slightly further away from the centre of the earth and consequently feels a smaller gravitational attraction. While a body at the poles is nearer to the centre of the earth and therefore experiences a greater gravitational force.

(ii)  Rotation of the earth

Because the Earth rotates, its gravitational pull on the body at the equator provides the body with a centripetal acceleration. This effect does not apply at the pole.

In other words, we say that: The value of the Acceleration due to gravity, g, depends on the distance from the centre of the earth to that place. The smaller the distance, the higher the value of g, and vice versa. Since the earth is not a perfect sphere, the distances from the centre of the earth to different points on the surface of the earth is different. Therefore the value of, g, varies from place to place. Thus, in places near to the centre of the earth like the north and south poles, the value of, g, is greater than at equator. It is also greater at sea level than at higher altitudes such as on top of mountains.

The acceleration due to gravity on the moon is about one-sixth of that of the earth.

(i.e. x 9.8 = 1.6 ms-2).

5.3  Mass and Weight of a Body

(a)  Mass

Mass is the quantity of matter in a body.

It is constant everywhere. I.e. does not vary from place to place.

SI unit of mass

The SI unit of mass is the kilogram (kg).

1. Weight (W)

Weight is the gravitational pull on a body by the earth. Or Weight is a measure of the pull of gravity on a body.

The direction of the weight of a body is always towards the centre of the earth.

SI unit of weight

Weight is a kind of force; therefore its SI unit is Newton.

Weight is calculated from the formula:

Weight  = Mass x Acceleration due to gravity

In symbols,   W  = mg

The weight of a body varies from place to place.

Explanation

From the formula W = mg, the weight of a body is directly proportional to the acceleration due to gravity. And since the g varies from place to place due to the variation in the distance from the centre of the earth, so does the weight of a body.

Note: Basing on the above explanation, it means that in places near to the centre of the earth like the North Pole and South Pole, the weight of a body is greater than when at the equator. It is also greater at sea level than at places higher than the sea level.

(c)  The Difference between Weight and Mass

 Mass Weight (i) Quantity of matter in a body – The pull of earth on a body (ii) Constant everywhere – Varies from place to place (iii) SI unit is kilogram (kg) – SI unit Newton (N) (iv) A scalar quantity – A vector quantity (v) Measured using beam balance or spring balance or electric balance calibrated in grams or kilograms – Measured using spring balance or electric balance calibrated in Newtons.

Worked Examples

1.  Calculate the weight of the following:

(a)  A box of mass 50 kg (b)  A boy of mass 2.5 kg

(c)  A bull of mass 200 kg (d)  A stone of mass 12 g

2.  Find the weight of an astronaut whose mass is 75 kg on:

(a)  The earth

(b)  The moon  (Take g = 10 ms-2 and g on the moon isth of g the earth)

3.  Find the masses of the weights:  (a)  400 N (b)  10 N

Solution

1. (a)  m = 50 kg, g = 10 ms-2, W = ? (b)  m = 2.5 kg, g = 10 ms-2, W = ?

W  = mg W  = mg

= 50 x 10 = 2.5 x 10

\W  = 500 N  \W  = 25 N

(c) m = 200 kg, g = 10 ms-2, W = ? (d) m = 12 g, = kg, g = 10 ms-2,W = ?

W  = mg W  = mg

= 200 x 10 = 0.012 x 10

\W  = 2000 N  \W  = 0.12 N

2. (a)  m = 75 kg, g = 10 ms-2, W = ? (b) m = 200 kg, g = 10 ms-2, W = ?

W  = mg W  = mg

= 75 x 10 = 75 x x 10

\W  = 750 N = 120 N

\W  = 125 N

3.  (a)  W = 400 N, g = 10 ms-2, m = ? (b)  W = 10 N, g = 10 ms-2, m = ?

W  = mg  W  = mg

400  = m x 10   10  = m x 10

m  = m  =

\ m  = 40 kg  \ m  = 1 kg

Self-Check 5.0

1.  What is the mass of a man on the earth if his mass on the moon 60kg.

A. 6kg B. 10kg C. 60kg D. 360kg

2.  Assume that you are taking measurements with a spring balance (dynamometer), where can you get the greatest reading for the same object?

A. At the centre of the earth B. On the moon

C. At the equator  D. At the poles.

3.  What is the name of any push or pull exerted?

A. Mass B. Force C. Friction D. Tension

4.  What do we call the pull of gravity on an object?

A. Mass B. Weight C. Moment D. Tension

5.  Which one of the following is not a measuring tool?

A. Equal-arm balance  B. Dynamometer  C. Lever D. Ruler

6.  Which one of the following is the unit of weight?

A. Newton B. Kilogram C. Meter D. Ton

7.  A mass of 60kg weighs 600N on the earth and 100N on the moon. What is the mass and weight of an object on the earth if it weighs 50N on the moon?

A. 60kg mass, 600N weight B. 10kg mass, 60N weight

C. 30kg mass, 300N weight D. 5 kg, mass, 100N weight

8.  Which one of the following is a force?

A. Energy B. Mass C. Weight D. Speed

9.  Which one of the following statements is not correct?

A. Force can change the speed of an object.

B. Force can change the shape of an object.

C. Force can change the direction of motion.

D. Force can change the mass of an object.

10.  Which one of the following are SI units of mass and weight?

A. g and n respectively B. N and kg respectively

C. kg  and g  respectively D. kg and N respectively

5.4  FRICTION

Friction is a force that opposes relative motion between two surfaces in contact with one another.

It results from two surfaces rubbing against each other or moving relative to one another.

It can hinder the motion of an object or prevent an object from moving at all.

The strength of frictional force depends on:

• The nature of the surfaces that are in contact,
• The magnitude of the force pushing them together and
• The weight of the object or objects.

In cases
involving fluid friction, the force depends upon

• The shape of the object moving through the fluid,
• Viscosity and
• Speed of an object as it moves through the fluid.

1. The Causes of Friction

Friction is thought to be caused by:

1. Small irregularities on the rough surfaces becoming interlocked.

E.g. rubbing two files together.

1. The attractive force between the surface molecules. This factor seems to play major role than interlocking of the small irregularities.

(b)  Types of friction

There are two types of friction, namely;

(i)  Static friction and

(ii)  Dynamic (Sliding or kinetic) friction.

1.  Static friction

Static friction is the frictional force between two surfaces that are not sliding over each other.

• Static friction occurs between stationary objects.
• It prevents an object from moving against a surface.
• It is the force that keeps an object e.g. a book from sliding off a desk, even when the desk is slightly tilted, and that allows you to pick up an object without the object slipping through your fingers.
• Static friction depends on:

– The coefficient of static friction (µs) between the surfaces of the objects

in contact,

– The nature of the surface and

– The normal reaction (R) of the object.

NB:  In order to move something, you must first overcome the force of static friction between the object and the surface on which it is resting.

Experiment 5.1  To Measure Static friction

Static friction can be measured using the apparatus shown in figure 5.1 below

• Place a wooden block on a table and connected it by a light string passing over a smooth pulley on to a mass hanger.
• Gradually increase force, F, by first adding 50 g and when the block is about to slide, continue adding small standard (known) 20 g or 10 g on to the mass hanger until the block just begins to slide or move.
• Read and record force, F, (the total mass on the mass hanger) at the point when the block just begins to slide.

Observations

At first the block remained at rest as the force is increased.

After some time, i.e. at certain value of the force on the pan, the block just begun to slip or slide in the direction shown on the diagram.

Explanation

When force, F, on the mass hanger was increased the frictional force that opposes the motion of the block also increases.

As more and more weights were added to the mass hanger, the frictional force reached its maximum value for the two surfaces in contact and begun to slide.

Result

The maximum value of the frictional force is equal to the total weight, F, on the mass hanger. This maximum frictional force is called limiting friction.

(b)  Limiting friction

Limiting friction is the force of friction between two surfaces when they are at the verge of sliding over each other.

The force of friction between an object and a surface can be calculated from the formula:

F = µR

Where:   F  = the force of friction,

µ  = the coefficient of friction between the object and the surface, and R  = the normal reaction.

(c)  Factors which determine the limiting friction

Limiting friction depends on:

(i)  The coefficient of static friction (µs) between the object,

(ii)  The nature of the surface and

1. The normal reaction (R) of the object.

These factors are demonstrated by attaching a spring balance calibrated in newtons to two blocks of wood of different weights as shown in the examples A and to the same block B below.

In example A, illustrates that limiting friction depends on weight. It shows that if it takes 10 N to slide a block of weight 50N on the floor, it will take 20 N of force to slide a block of weight twice that of 50 N (i.e.100 N).

While example B illustrates that friction is independent of the area in contact. That is it does not depend on the size of surface area in contact between the objects. It shows that the limiting friction required to move 250 N for the different surface areas.

(d)  Coefficient of static friction

Coefficient
of static friction is the ratio of limiting friction to the normal reaction between surfaces. It is denoted by a symbol, m, pronounced mew.

Experiment  5.2  To determine the value of coefficient of friction, m

Apparatus:  A wooden block, a string, mass hanger, heavy standard weight (masses), slotted masses of 50 g and 10 g, a smooth pulley and a weighing balance.

Procedure

Arrange the apparatus as shown in figure 5.3 below.

• Weigh the block of wood and record its mass, Mb, in kg.
• Gradually increase force, F, by first adding 50 g and when the block is about to slide, continue adding small standard (known) 20 g or 10 g on to the mass hanger until the block just begins to slide or move.
• Record the total mass, m, hence the force F, on the mass hanger.

• Load the block of wood by adding a heavy standard mass M = 1 kg.
• Repeat the procedure (b) to (c).
• Now repeat the procedures (b) to (d) for values of M = 2, 3, 4, 5 and 6 kg.
• Enter your results in the table 5.1 below.

Table of results

 (Mb + M)/kg W =(Mb + M)g = R /N m/kg F = mg/N Mb + 0 Mb + 1 Mb + 2 Mb + 3 Mb + 4 Mb + 5 Mb + 6

Table  5.1

• Plot a graph of F against R (normal reaction = W).
• Find the slope of the graph.

Note that the value of F is equal to the total weight on the mass hanger which is equal to the limiting friction for each total weight.

If the experiment is accurately performed, the graph in the figure below is obtained.

The graph of limiting friction (F) against normal reaction

The slope of the graph is called the coefficient
of friction, m, and is given by the formula.

Coefficient of static friction, =

ms
=

\ Limiting friction, F  = μR

Note that:

• Since normal reaction R, is always greater than the limiting friction F,

i.e. R > F, the value of m is always less than 1.

Worked Examples

Example 1

A block of wood of mass 5 kg is placed on a table top. Find the limiting friction if the coefficient of friction is 0.5. (Take g = 10 m/s2).

Solution

Data:  μ = 0.5, m = 5 kg, g = 10 m/s2, Limiting friction, F = ?, R = ?

Hint:  Limiting friction is calculated from the formula F = μR. But R is not given in the data. This should not make the whole thing difficult to you.

Always remember that R = W. So using the formula W = mg, find the value of W and then finally use the formula F  = μR to
get F as shown below.

First calculate W: W  = mg

= 5 x 10

= 50 N

From Weight  = Reaction  \ R = 50 N

Now that you have known R, you can now use the formula;

Limiting friction, F = μR

= 0.5 x 50

\ F  = 25 N

Self-check

1.  A chalk box of mass 400 g is placed on a table. Find the limiting friction if

μ = 0.2, (g = 10 ms-2). Answer:  0.8 N

2.  A block of wood of mass 2 kg is placed on a table. Find the limiting friction if

μ = 0.4, (g = 10 ms-2). Answer:  8 N

2.  Kinetic or Sliding friction

The kinetic or sliding friction is the frictional force between any two surfaces that are moving relative to each other.

• It resists the motion of an object as it moves along a surface.
• Sliding friction occurs between objects as they slide against each other.

(b)  Factors on which Sliding friction depends

Like the limiting friction, sliding friction depends on:

• The weight of the sliding object,
• The nature of the surfaces in contact and
• The coefficient of kinetic friction.

Experiment 5.3  To Measure sliding friction

Sliding or kinetic friction can be measured using the apparatus shown in figure 5.4 below

• Place a wooden block on a table and connected it by a light string passing over a smooth pulley on to a mass hanger.
• Gradually increase force, F, by first adding 50 g and 20 g or 10 g on to the mass hanger until the block is about to slide.
• Give the block a push.
• Read and record force, F, (the total mass on the mass hanger).

Result

The maximum value of the frictional force is equal to the total weight, F, on the mass hanger. This maximum frictional force is called sliding or kinetic friction, Fk.

Note that:

1.  The limiting friction is always greater than the sliding friction.

1. The procedure for the measurement of kinetic friction and the coefficient of kinetic friction, mk, are the same as for static friction. The only difference is that a push is required for the determination of the kinetic friction at the point when the body is about to slide.

5.6  The angle of friction

When a block of wood is placed on an inclined plane at an angle of friction α, such that the block is just on the point of slipping as shown in the diagram below.

Figure 5.5

From Coefficient of friction, m  =

• =

F  = mR    ……………………………..  (1)

m
= But = Tan a

\m  = Tan a  ……………………………..  (2)

Note:  That the tangent of the angle of friction is equal to the coefficient of friction.

Substituting equation (1) in equation (2) we have,

F  = TanaR

Note:  Do not worry if you do not understand the steps in the derivation of the above formula. You will understand it when you cover trigonometry in mathematics in S.2.

5.7  (a)  Laws of friction

(i)  Friction always opposes motion

(ii)  Friction depends on the nature of the surface in contact.

(iii)  Friction – is much less on smooth surface and is much greater on rough

surface.

(iv)  The limiting friction and sliding friction are both proportional to the normal reaction i.e. F µ R

Friction is useful in our daily life. This is seen in the following:

1.  Motion

Friction allows motion of objects, as seen in the following:

1. Walking Friction between the foot and the ground allow a person to walk.

In a slippery place, there is no friction. The result is that

movement becomes difficult.

2. Clutch Friction between clutch plates allows the engine to drive the

wheels.

3. Tyres   Friction between the tyres and the road surface allows the torque

of the wheels to push cars or bicycles along the road.

4. Brakes   Friction between the brake lining and the brake drum halts the

vehicle.

Friction between the foot of the ladder and the ground prevents it from slipping off.

3.  Lubrication

Liquid friction helps to spread oil round an axle.

(i)  Friction between the moving parts of machines reduces the efficiency of machines and often causes much of the work done by the effort to be wasted.

(ii)  It causes surfaces to wear off. E.g. tyres and soles of shoes.

(iii) It also causes unnecessary noise and heat.

Where friction becomes a disaster, we can reduce it by applying some methods.

5.8  Ways of reducing friction

1.  Lubrication

In lubrication, oil or grease is introduced between surfaces that sliding over one another. The oil film keeps rough surfaces apart, thus preventing them from coming into direct contact.

2.  Ball bearings

Bearing is a mechanical device for decreasing friction in a machine in which a moving part bears—that is, slides or rolls while exerting force—on another part.

Steel ball bearings are used for example in an axle and shaft of the bicycle then the axle and shaft. When the ball bearings are used, the axle and shaft do not slide but roll over one another. Rolling friction is less than the sliding friction; therefore friction is lessened by means of ball bearing.

3.  Rollers Friction can be reduced by placing rollers between the two surfaces in contact.

4.  Smoothening of surfaces in contact

Friction can also be reduced by making rough surfaces smooth where necessary.

5.  Streamlining  Streamlining refers to shaping of an object such that the layer of air easily slides off. Notable examples are seen in:

– aeroplane, submarines, boats, fish, birds etc.

5.9  Fluid Friction

Fluid Friction
is a retarding force which acts on a body moving through a fluid.

It is due to the internal friction existing between layers of fluids of a liquid in motion. It is caused by the attraction between the molecules of one layer and the molecules of another layer. The frictional resistance of a fluid (liquid or gas) opposing motion of a body moving through it is called viscosity.

Factors which determine the magnitude of fluid friction

1.  Unlike friction between solids’ viscosity is proportional to the surface area and the velocity of the object moving through the fluid.

2.  Nature of fluid. Viscosity depends on the type of fluid for example glycerin, engine oil, and natural honey flow much less easily from a vessel than liquids such as petrol, paraffin and water.

3.  Temperature:  Viscosity of liquids decreases as a temperature of liquid increases while that of gases increases with temperature.

NB:  Liquids which have high viscosity e.g. glycerin, engine oil are called vicious liquids and are used to make lubricants.

5.10  Scalar and Vector Quantities

(a)   Scalar Quantity

A scalar quantity is a quantity which has only magnitude. They only give the amount of things. Examples scalar quantities are:

– Mass  e.g.  5 kg of sugar

– Time 2 hours

– Volume 4 litres

– Density 0.8g/cm3

– Temperature  50 ºC

(b)  Vector Quantity

A vector quantity is a quantity which has both magnitude and direction. Examples of vector quantities are:  – Force or weight,

– Velocity,

– Acceleration,

– Momentum.

5.11  Representation of vectors

Vectors are represented by an arrow headed line drawn to scale is called a vector. A vector has the following properties:

• An application point (where it is applied).
• A magnitude (size i.e. how big or small it is).
• A direction.

When two or more forces act on a body their combined effect called, resultant force. The resultant
force
on the body depends on their relative directions, (i.e. the angle, q, between the forces).

Resultant force

Resultant force is a single force which has an effect of two or more forces acting at the same point.

The forces which form a resultant force are called component forces.

Resultant force is found by using the following equation:

Resultant force  = Sum of forces

I.e    = + + +

Points to note: Since force is a vector, the direction is important. Therefore,

• If force(s) to the right are taken to be positive, then the one(s) to the left is/are negative. Like wise,
• If force(s) acting to north or east is/are taken to be positive, then the one(s) acting to south and west respectively is/are taken to be negative.
• The number of Fs to the right of the above formula = to the number of forces acting on a body.

(a) Forces acting in a straight line

The resultant force due to forces acting in a straight line is got by addition or subtraction of the forces depending on their directions.

(i)  Two forces acting on a straight in the same direction (q = 0°)

Consider two forces F1 and F2 acting on body P as shown in figure below.

Since the forces are acting in the same direction, they are taken to be positive.

The resultant force, R, is got by addition of the forces.

i.e  = +

Example 1

Two forces 20 N and 50 N act on body R in the same direction as shown in the diagram below. Find the resultant force.

Data: F1 = 20 N, F2 = 50 N , = ?

= +

= 20 N + 50 N

\  Fr  = 70 N

(ii)  Two forces acting on a straight in opposite directions (q = 180°)

Consider two forces F1 and F2 acting on body M in opposite directions as shown in figure 4.8 below.

Since the forces are acting in opposite directions, one is taken to be positive and the other negative.

Let F1 be positive and F2 be negative.

The resultant force, given by using the equation:

= +

= +

\  =

Therefore, the resultant force, ,  = () in the direction of F1.

Example 2

Two forces 300 N and 50 N are acting on a body Q as shown in figure 4.9 below. Find the resultant force.

Solution

F1 = 300 N, F2 = 50 N

Since the forces are acting in opposite directions, one is taken to be positive and the other negative.

Let 300 N be positive and 50 N be negative.

The resultant force, given by using the equation:

= +

= 300 + -50

= 300 – 50

\ Fr  = 250 N

Therefore the resultant force is 250 N in the direction of 300 N (to the right).

NB:  Instead of taking one force to be positive and the other negative, you may simply subtract the smaller force from the larger one.

1. More than two forces acting on a straight in opposite directions (q = 180°)

Consider four forces F1, F2, F3 and F4 acting on body S as shown in the diagram below. Find the resultant force.

Solution

For the case such as above where there are more than two forces acting on a body in opposite direction, follow the following steps.

Step I: Get the resultant forces in the directions shown in the diagram above.

Resultant force towards right, Rr   = (F1 + F2) N

Resultant force towards left, Rl = (F3 + F4) N

Step II: Get the final resultant force, R, of the two resultant forces  above.

Resultant force, R  = Rr – Rl

= N

Example 3

Four forces are acting on body V as shown in the diagram below. Find the resultant force.

Solution

Resultant force acting towards right  Rr  = (F1 + F2) N

= 80 N + 120 N

= 200 N

Resultant force acting towards left  Rl
= (F3 + F4) N

= 100 N + 50 N

= 150 N

Resultant force, R, acting on V = Rr – Rl

= 200 N – 150 N

= 50 N acting towards right.

(b)  Parallel forces

The resultant vector for two or more vectors can be represented by a head-to-tail addition of arrows. For example if two or more forces act in the same direction, the resultant force is obtained as shown below.

NB:  The little arrow over F indicates that we are dealing with vector (a quantity for which direction is important).

c)  Forces acting at right angles to each other (q = 90°)

When two forces act on a body at an angle to each other, the resultant force can be obtained by applying any of the following laws.

• The triangle law of forces and
• The parallelogram law of forces.

(i)  Triangle law of forces

The law states that “if a body is acted upon by two forces that are at an angle to each other, the resultant of the forces is the third line drawn to complete the triangle”.

For example, if two forces of equal or unequal magnitude, one directed due east and the other directed due north act on an object at rest but free to move in the direction of the resultant force, the magnitude of the resultant force is shown by the length of the arrow representing the resultant force.

Consider the diagram in figure 5.9 below in which two forces F1 and F2 act on a body. F1 acts due north and F2 act due east as shown below.

Using the method of head-to-tail addition of vectors at right angle, the right angle triangle is completed by using the tail of F1 on the head of
F2 as shown in the diagrams below and the resultant force, Fr, is got by applying Pythagoras theorem as shown in the diagrams below.

Worked Examples

1.  Two forces 30 N and 40 N act on a body of mass 20 kg at right angle to each other. Calculate;  (a)  The resultant force

(b)  The acceleration of the body.

Solution:  (a)  A sketch diagram showing the forces

Applying the Triangle law of forces and head-to-tail addition of vectors, we complete the diagram as below.

Using Pythagoras theorem

F2   = 402 + 302

=

=

=

\ F  = 50 N

(b)  F = 50 N, m = 20 kg, a = ?

F  = ma Þ
a = =
= 2.5 ms2

2.  The diagram in the figure shows 3 forces of 8N, 6N and 16N acting on a particle X. Find the resultant force magnitude of the resultant force is

Solution

Step I  First get the resultant force between 8N and 16N, since they are on a straight line.

Resultant force = 16 N – 8 N

= 8N  acting to left (the direction of the larger force).

Step II  Join the two forces using a head to tail and complete it to form a right angle triangle. Then find the hypogenous by using Pythagoras Theorem as in example 1 above.

Using Pythagoras theorem

Fr2  = 62 + 82

Fr  =

=

=

\ Fr   = 10 N

(ii)   The parallelogram law of forces.

If two forces, acting at one point on the same object, are arranged in magnitude and direction by the sides of a parallelogram drawn from the point, their resultant is represented in both magnitude and direction by the diagonal of the parallelogram drawn from the point.

In this method, the vectors to be added are represented by arrows joined tail-to-tail instead of head-to tail as before. The resultant is obtained by completing the diagonal of the parallelogram as shown in figure 5.10 below.

Composition and Resolution of Vectors

Composition of vectors is the substitution of two or more vectors with a single vector which has the same effect as all the separate vectors combined.

Resolution of vectors is the separation of a single vector into two vectors acting in definite directions upon the same point. Each of the two vectors is called a component of the single vector. One is the horizontal component and the other is the vertical component.

Self-Check Questions 5.1

1.  Force is given by the product of

A. displacement and velocity B. displacement and mass

C. acceleration and mass   D. velocity and mass

2.  Forces of 60N, 10N, 40N and 10N act on a body as shown in figure 5.11. In which direction does the body move?

A. upwards B. downwards

C. to the left   D. to the right

3.  Two forces of 5N and 12N act on a body at right angles. Find their resultant

A. 7 N   B. 13 N    C. 17 N D. 169 N

4.  A Newton is defined as the

A. unit of force.

B. force which produces an acceleration of 1ms-2

C. force which gives a mass of 1 kg an acceleration of 1ms-2

D. force which gives any mass an acceleration of 1 ms-2

5.  Two forces of 3N and 4N act at point at right angles to each other. The magnitude of their resultant force is

A. 25N B. 7N C. 5N D. 1N

6.  In the diagram in the figure, the magnitude of the resultant force acting on a body P is

A. 1N

B. 5N

C. 7N

D.11N

7.  A force of 1N acts on a mass of 0.05kg initially at rest. Its acceleration is

A. 0.05 ms-2.  B. 2 ms-2. C. 0.5 ms-2. D. 20 ms-2.

8.  Find the force required to give a mass of 500g an acceleration of 2 x 10-2 ms-2.

A. 1 x 10-2 N B. 1 x 10-1 N C. 1 x 102 N D. 1 x 104 N.

9.  In which of the following situations is minimum friction force required.

A. Sliding down a slope B. Walking along a road.

C. Leaning a ladder against a wall D. Designing brake blocks for bicycle.

10.  Which one of the following are true statements about friction?

(i) It does not oppose friction.

(ii) It causes wearing of surface.

(iii) It decreases as weight of a body decreases.

(iv) It can be reduced by applying oil between surfaces.

A. (i) only. B. (i) and (iii) only. C. (ii), (iii) and (iv).  D. All

11.  Which of the following physical properties changes when a body is moved from the earth to Pluto (outermost planet of the solar system)?

A. mass B. volume C. weight D, density

12.  Which of the following group consists of vectors only

A. momentum, acceleration, time, energy

B. speed, velocity, displacement, energy

C. displacement, velocity, acceleration, force

D. velocity, work, power, momentum

13.  A force of 10N acts on a body and produces an acceleration of 2ms-2. The mass of the body is

A. 0.2 kg B. 5.0 kg C. 20.0kg D. 50.0kg

14.  A stone of mass 2.5g is thrown with an average force of 5.0N. Find the average acceleration in m/s2.

A. 5.0 x 10-4  B. 2.0 x 10-3  C.25 x 102  D. 2.0 x 103

15.  Two forces act on chest, one of them is 50N and due northeast, and the other one is 75N and due south west. What is the resultant force?

A. 125 N due Northwest B. 125N due southwest

C. 25N due North West D. 25 N due south west

16.  Which of the following sets contains only vector quantities?

A. Weight, displacement, acceleration, magnetic field.

B. Energy, electric field, momentum, distance.

C. Mass, velocity, force, speed.

D. Specific heat capacity, power, time, volume.

17.  A block of mass 10 kg accelerates uniformly at a rate of 3 ms-2 along a horizontal table when a force of 40N acts on it. Find the frictional force between the block and the table

A. 10N B. 13.3N C. 30N D. 70N

18.  Which one of the following quantities is the odd one out?

A. force. B. velocity. C. momentum.  D. kinetic energy.

19.  Find the force that acts on a body of mass 0.05kg accelerating at 20 m/s2

A.1.0N B.10N C.100N D.400N

20.  A box is pulled horizontally with a force of 100N inclined at an angle 50° to the horizontal as shown in the figure 5.12. The effective force, in Newton, is

A. 100/sin

B. 100sin 500

C. 100/cos500

D. 100cos500

21.  A block of mass 8kg slides on a rough horizontal surface under the action of a force of 48N as shown in the figure below. If the block moves with an acceleration of 5 ms-2, calculate the frictional force on the block

A. 6N

B.8N

C. 40N

D. 48N

22.  What is the force that slows down or stops sliding?

A. Weight B. Gravity C. Friction D. Tension

23.  A force of 20N acts on an object and there is 5N frictional force between the object and the surface. What is the net force on the object and its direction?

A. 15 N in the direction of 20 N  B. 15 N in the direction of the frictional force.

C. 25 N in the direction of 20 N  D 25 N in the direction of frictional force.

24.  Which one of the following is a vector quantity?

A. Force B. Time C. Temperature.  D. Mass

25.  Three forces 25N, 30N, 40N act on object on the same line either in the same direction or in opposite directions. Which one of the following can not be the resultant of these forces?

A. 5N B. 15N C. 35N D. 95

SECTION B

(Where necessary take g = 10 ms-2)

26.  (a)  Distinguish between weight and mass of a body.

(b)  The force of gravity on the moon is one-sixth of that on the earth. Determine the weight of a 12 kg mass on:

(i)  The earth, (ii)  The moon?

(c)  Explain why the weight of a body on earth may vary from place to place.

27.  (a)  (i) What is meant by dynamic friction?

(ii) Describe, with the aid of a diagram, an experiment to determine the limiting friction between two surfaces in contact.

(iii)  Sate the any two factors, which affect friction.

(b)  Give two applications of friction.

28.  (a)  What is a vector quantity?

(b)  Two forces of 30N and 40N act perpendicularly on an object of mass 10 kg as shown in the figure. Calculate:

(i)  The magnitude of the resultant force on the object.

(ii)  The acceleration of the object.

1. The figure shows the resultant R of two forces P and Q.

R makes an angle of 600 with a horizontal and P is 50N.   Find the magnitude of:

1. Q
2. R

(d)  Two forces of 6N and 10N act at the same time on a body P of mass 500g as shown in the figure 5.15. Find the:

(i) Resultant force on P.

(ii) Acceleration of P.

29.  A body of mass 5 kg is acted upon by two forces of 3 N and 4 N at right angles to one another. Find:  (a)  By calculation the resultant of the two forces.

(b)  The acceleration of the body.

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