The document discusses different types of forces including contact forces, non-contact forces, frictional forces, and buoyant forces. It defines force as a push or pull that can change the motion of an object. There are two main types of forces - contact forces that act directly between objects in contact, and non-contact forces that act over a distance through force fields. Examples of both types are given. Frictional forces oppose the relative motion of objects in contact and can be useful or disadvantageous depending on the situation. The coefficient of friction and methods for measuring it are also described.

1. By Zeba Mubarak
Twinikli International Senior High School

2. RECAP (FORCES)
FORCE Is defined based on what it can cause bodies to do, it may cause a
body’s length to increase or decrease. IT CAN CAUSE A BODY’S REST
POSITION TO CHANGE OR IF IN MOTION TO CHANGE IT DIRECTION
hence
FORCE is defined as a push or a pull that can change a bodies
state of rest or uniform motion in a straight line.

3. TYPES OF FORCES
•THERE ARE TWO TYPES OF FORCES. THESE ARE
CONTACT FORCE / LOCAL FORCE
NON – CONTACT / NON – LOCAL / FORCE FIELDS

4. CONTACT FORCE
•They are forces that are experienced by bodies when they
are in direct contact with the source of the force.
•EXAMPLES
Upthrust
Frictional force
Tension force
Surface tension
Forces exerted on a ball when kicked
Viscous force (fluid resistance)

5. NON –CONTACT FORCE
•They are forces that are experienced by bodies that may or
may not be in direct contact with the source of the force.
•EXAMPLES
Magnetic force
Gravitational force
Electric Force
Nuclear force
NB: CONTACT FORCE ARE CALLED LOCAL FORCE BECAUSE THE SOURCE OF THE FORCE MUST BE IN THE
AREA(LOCALITY) OF THE OBJECT.

6. FRICTIONAL FORCE•It is the tangential force that acts on surfaces in contact and which opposes
their relative motion. It is experienced by solid bodies in contact.
•USES / ADVANTAGES OF FRICTIONAL FORCES
It makes body in motion to stop
It is use for sharpening cutlasses
It makes walking possible
It enables us light fire
It makes writing possible
It enables a screw or a nail to remain in place after being screwed into
position.

7. •DISADVANTAGES OF FRICTIONAL
 It makes the sole of shoes wear and tear
It produces heat when 2 solid bodies in contact makes relative motion
It reduces the efficiency of a machine
•HOW WE WILL REDUCE FRICTION
 Greasing solid surfaces in contact
Introducing impurities between surfaces reduces friction
By introducing spherical metallic balls in between two metals moving
over each other has in ball bearings or race reduces friction.

8. BALL RACE OR BEARING & LABELLING
Spherical body
Metal run
Metal

9. TWO TYPES OF FRICTION
•THESE ARE:
1) STATIC – These opposes motion of bodies when they are stationary
2) DYNAMIC - These opposes motion of bodies already in motion
Static friction is always bigger than they the dynamic frictional force

10. NORMAL FORCE OR REACTION
It is the component of a supported force that is perpendicular to
the supporting surface.
HORIZONTAL SURFACE

11. •EXAMPLES
Determine the reaction exerted on a surface when a 20 kg body
is placed on the surface. Assume horizontal.
SOLUTION
Reaction = 𝑚 × 𝑔
= 20 × 10
= 200 𝑁

12. INCLINED SURFACE
𝑚𝑔𝑐𝑜𝑠 𝜃 causes the body to remain on the surface
𝑚𝑔𝑠𝑖𝑛 𝜃 causes the body to pulled down the surface

13. •EXAMPLES
A 10 kg mass rest on a surface at 10° to the horizontal. Calculate
i. The force that presses the body unto the plane
ii. The force that tries to pull the body down along the surface
iii. The reaction

14. 1) 𝑚𝑔𝑐𝑜𝑠 𝜃 = 10 × 10 cos 10 = 98.48 𝑁
2) 𝑚𝑔𝑠𝑖𝑛 𝜃 = 10 × 10 sin 10 = 17.36 𝑁
3) 𝑅 = 𝑚𝑔𝑐𝑜𝑠 𝜃 = 10 × 10 cos 10 = 98.48 𝑁
LIMITING STATIC FRICTION
This is the minimum force required to move a body at rest. The
force is described as dynamic limiting frictional force when it is
the minimum force that must be applied on a moving body to make
the body move at a constant velocity.

15. •Coefficient of Static Friction
It is defined as the
𝐥𝐢𝐦𝐢𝐭𝐢𝐧𝐠 𝐬𝐭𝐚𝐭𝐢𝐜 𝐟𝐫𝐢𝐜𝐭𝐢𝐨𝐧𝐚𝐥 𝐟𝐨𝐫𝐜𝐞
The normal force or reaction
.
𝜇 =
𝐹𝑥
𝑅
or 𝐹𝑥 = 𝜇 × 𝑅
Example
A 15 kg body is on a horizontal surface which has a coefficient of
friction of 0.25. Calculate
i) The normal force
ii) The limiting static frictional force

16. • Solution
i. The normal force = 𝑚 × 𝑔 = 15 × 10 = 150 𝑁
ii. The limiting static frictional force = 𝐹𝑥 = 𝜇𝑅 = 0.25 × 150 = 37.5 𝑁
EXAMPLE
A 100 kg body is on a plane inclined 30 ° to the horizontal, if the frictional force on
the body is 50.0 N, calculate the
i. The normal force
ii. The coefficient of friction

17. •Solution
i. The normal force = 𝑚𝑔𝑐𝑜𝑠 𝜃 = 100 × 10 cos 30 = 866.03 𝑁
ii. The coefficient of friction 𝜇 =
𝐹𝑥
𝑅
=
50 𝑁
866.03
= 0.058 𝑁
RESULTANT FORCE – This is the difference between the total force in the
direction of motion and the total force opposite the direction of motion
𝑹 𝒇 =Total force in the direction of motion – total force opposite the direction
of motion
Or it can also be determined by mass × acceleration = 𝒎 × 𝒂 = 𝒎𝒂

18. EXAMPLE
A 20 kg body on a horizontal surfaces is pulled to the right with a force of
100N. Determine.
i. The frictional force
ii. The resultant force
iii. The acceleration 𝑇𝑎𝑘𝑒 𝜇 = 0.2

19. • SOLUTION
• I. Frictional force , 𝐹𝑥 = 𝜇 × 𝑅 , 𝑅 = 𝑚𝑔 = 20 × 10 = 200 𝑁
𝐹𝑥 = 0.2 × 200 = 40.0 𝑁
II. Resultant force , 𝑅𝑓 =Total force in the direction of motion – total force opposite the
direction of motion = 100 − 40 = 𝟔𝟎 𝑵

20. III. Acceleration
𝑹 𝒇 = 𝒎𝒂𝒔𝒔 × 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏
60 = 20 × 𝑎
∴ 𝑎 =
60
20
= 𝟑 𝒎𝒔−𝟐
Home work 1
A 100 kg toy car is pulled up a plane inclined 30 ° to the horizontal with a force of
1000 N. Given that the 𝐹𝑥 coefficient of the force is 0.25,
Calculate
i. 𝐹𝑥 force ii. Total force down along the plane iii. The resultant force
iii. The acceleration of the body act along the plane

22. • SOLUTION
i. FRICTIONAL FORCE, 𝐹𝑥 = 𝜇 × 𝑟 , 𝑟 = 𝑚𝑔𝑐𝑜𝑠 𝜃
𝐹𝑥 = 𝜇 × 𝑚𝑔𝑐𝑜𝑠 𝜃 = 0.25 × 100 COS 30 = 𝟐𝟏𝟔. 𝟓 𝑵
ii. TOTAL FORCE DOWN ALONG THE PLANE = 𝑚𝑔𝑠𝑖𝑛 𝜃 + 𝐹𝑥 = 100 × 10 𝑠𝑖𝑛30 + 216.5
= 716.5 𝑁
iii. RESULTANT FORCE, 𝑅𝑓 = 𝑡𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑡𝑖𝑜𝑛 −
𝑡𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛 = 1000 − 716.5 = 𝟐𝟖𝟑. 𝟓𝑵
iv. ACCELERATION =
𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒
𝑚𝑎𝑠𝑠
=
283.5
100
= 2.835 𝑚𝑠−2

23. Question
A 200 kg body rest on a surface inclined 25 ° to the horizontal. If this body
is pulled down along the plane with a force of 100 N. Calculate
i. 𝐹𝑥 force ii. Total force in the direction of motion iii. The
resultant force iii. The acceleration of the body act along the plane
[Take 𝜇 𝑎𝑠 0.3]

25. • SOLUTION
i. FRICTIONAL FORCE, 𝐹𝑥 = 0.3 × 𝑟 , 𝑟 = 𝑚𝑔𝑐𝑜𝑠 𝜃
𝐹𝑥 = 𝜇 × 𝑚𝑔𝑐𝑜𝑠 𝜃 = 0.3 × 2000 COS 25 = 𝟓𝟒𝟑. 𝟕𝟖𝑵
ii. TOTAL FORCE IN THE DIRECTION OF MOTION= 𝑚𝑔𝑠𝑖𝑛 𝜃 + 𝐹𝑑 = 200 × 10 𝑠𝑖𝑛25 + 100
= 𝟗𝟒𝟓. 𝟐𝟒 𝑵
iii. RESULTANT FORCE, 𝑅𝑓 = 𝑡𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑡𝑖𝑜𝑛 −
𝑡𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛 = 945. 24 − 543.78 = 𝟒𝟎𝟏. 𝟒𝟔𝑵
iv. ACCELERATION =
𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒
𝑚𝑎𝑠𝑠
=
401.46
200
= 𝟐. 𝟎𝟎𝟕 𝒎𝒔−𝟐

26. MEASUREMENT OF COEFFICIENT OF 𝐹𝑥
The test body is placed on a horizontal table that has a pulley fixed to one end of a thread that
is passing over a pulley attached to a scale pan. Masses are gently placed on the scale pan
until the test body just begins to move.
Weigh the mass on the scale pan and record it as 𝑀𝑠. WEIGH THE TEST BODY AND RECORD THE MASS
AS 𝑀
THEORY
WHEN THE BODY JUST BEGINS TO MOVE 𝑀𝑠 𝑔 = 𝑇
BUT 𝑇 = 𝑓𝑠 HENCE 𝐹𝑥 = 𝑚 𝑠 𝑔 REACTION 𝑅 = 𝑀𝑔
∴ COEFFICIENT OF FRICTION 𝜇 =
𝐹𝑥
𝑅
=
𝑀𝑠 𝑔
𝑀𝑔
=
𝑀𝑠
𝑀

28. TO DETERMINE 𝜇 USING INCLINED PLANE
• The test body is placed on a plain. One end of the plane is gradually raised until the body
just begins to move.
• The angle of inclination 𝜃 at this instant is measured with a protector. 𝛍 = 𝐭𝐚𝐧 𝛉

29. When a body just begins to move 𝐹𝑥 =
𝑚𝑔 sin 𝜃
But 𝐹𝑥 = 𝜇𝑅
𝑅 = 𝑚𝑔𝑐𝑜𝑠𝜃
𝜇𝑚𝑔𝑐𝑜𝑠 𝜃 = 𝑚𝑔𝑠𝑖𝑛 𝜃
𝝁 =
𝑚𝑔𝑠𝑖𝑛 𝜃
𝑚𝑔𝑐𝑜𝑠 𝜃
=
sin 𝜃
cos 𝜃
= 𝐭𝐚𝐧 𝜽

30. ARCHIMEDES PRINCIPLE
• A body in a fluid apart from experiencing its own weight also experiences a vertically
directed upward force that tends to reduce the weight of the body.
• Hence, bodies in fluid weigh less. UPTHRUST OR BUOYANT FORCE is the vertically directed
force experienced by bodies in fluid.
• The volume of fluid displaced = the volume of the part of solid or body immerse in
fluid
• The weight of the fluid displaced = the volume of solid immersed × density of fluid ×
gravity
• The weight of fluid displaced = UPTHRUST.
• The apparent loss in weight of bodies is the UPTHRUST

31. QUESTIONS
• A piece of wood has a mass of 200 g. When placed in 𝐻2 𝑂 with 50 𝑐𝑚3 of the wood in
𝐻2 𝑂 , there is a loss in mass. Calculate
• Volume of 𝐻2 𝑂 displaced
• Mass of 𝐻2 𝑂 displaced
• The upthrust on the wood
• Find the apparent loss in mass of the wood.

32. SOLUTION
1. Volume of 𝐻2 𝑂 displaced = 50 𝑐𝑚3
2. Mass of 𝐻2 𝑂 displaced = Density × Vol. of the 𝐻2 𝑂 = 1 × 50 = 50 g
3. The upthrust on the wood =
50
1000
= 0.050 × 10 𝑚/𝑠 = 0.5 𝑁
4. Apparent loss in mass of the wood = Mass in air – mass in fluid = 200 − 50 = 150 𝑔

33. A 20 𝑐𝑚3
balloon is left in air of density 0.0014 𝑔/𝑐𝑚3
. Calculate the mass of air displaced and hence, the
upthrust.
Answer
Density =
𝑀𝑎𝑠𝑠
𝑉𝑜𝑙𝑢𝑚𝑒
Mass = Density × Volume = 0.0014 × 20 = 0.028 g
Upthrust =
0.028𝑔
1000
= 0.000028 × 10 = 𝟎. 𝟎𝟎𝟎𝟐𝟖 𝑵
Archimedes Principle
When a body is fully or partially immersed in a fluid, it experiences an upthrust equal to the weight of fluid
displaced.
NB: The upthrust reduces the weight.

34. FLOATATION
•Law of Floatation
A floating body displaces its own weight of fluid in the fluid in which its floates.
During Floatation
• The volume of fluid displaced = the volume of the body immersed.
• Mass of fluid displaced = Mass of the body
• Upthrust = weight of body

36. METHOD
• 1. Fill an overflow can with 𝐻2 𝑂 upto the spout level
• 2. Weigh an empty beaker with an electric balance to record the mass 𝑀1
• 3. Place the beaker below the spout
• 4. Place the test tube in the 𝐻2 𝑂 and add lead shots to the tube for the tube to float upright.
• 5. Weight the beaker with the overflow 𝐻2 𝑂 and record the mass 𝑀2
• 6. Evaluate the mass of the overflow 𝐻2 𝑂 AS 𝑊 = 𝑀2 − 𝑀1
• 7. Remove the test tube with the leadshots from the 𝐻2 𝑂, weigh and record the mass as 𝑀

37. OBSERVATION AND CONCLUSION
• It is found that 𝑀 = 𝑚𝑔 indicating that mass of the test tube/ body is equal to the mass of
overflow water
• Weight of body = 𝑚𝑔
• Upthrust = mass of fluid × g = 𝐴ℎ 𝜌 𝑓𝑙𝑢𝑖𝑑 𝑔
• 𝐴 = 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎, ℎ = ℎ𝑒𝑖𝑔ℎ𝑡 , 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝐴ℎ = 𝑉𝑜𝑙𝑢𝑚𝑒
𝑚𝑔 = 𝐴ℎ𝜌𝑔
𝑚 = 𝐴ℎ𝜌(𝑑𝑒𝑛𝑠𝑖𝑡𝑦)
𝑚 = 𝑉𝜌

38. QUESTIONS
• A test tube has a mass of 15 g, a cross sectional area of 1.5 𝑐𝑚2
. The test tube floats
in the liquid of 𝜌 = 0.8 𝑔/𝑐𝑚3. Calculate the depth of immersion of the test tube.
SOLUTION
𝑀 = 15 𝑔 𝐴 = 1.5 𝑐𝑚2 𝜌 = 0.8 𝑔/𝑐𝑚3
𝑀 = 𝐴ℎ𝜌
15 = 1.5 × ℎ × 0.8
ℎ =
15
1.2
= 12.5 𝑐𝑚

39. QUESTION
• A piece of wood floats in a liquid of relative density(RD) = 1.02 if the mass of the
wood is 25 kg and the volume of wood above the liquid is 0.0071 𝑚3. Calculate the
total volume of the wood
SOLUTION

40. 𝑅. 𝐷 = 1.02 𝜌 = 𝑅. 𝐷 × 𝜌 𝑤 = 1. 02 × 1000 = 1020 𝑘𝑔𝑚−3
𝑣𝑜𝑙 𝑜𝑓 𝑤𝑜𝑜𝑑 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒 𝑙𝑖𝑞𝑢𝑖𝑑 = 0.0017, 𝑀 = 25 𝑘𝑔,
𝑀 = 𝑉2 𝜌
𝑉2 =
25
1020
= 0.0245 𝑚3
Volume of wood = 𝑉1 + 𝑉2 = 0.0071 + 0.0245 = 0.0316 𝑚3

41. QUESTIONS
• A 200g wood float in 𝐻2 𝑂. If the top part of the wood is just above covered water. What
is the size length of the wood
SOLUTION
𝑀 = 𝑉𝜌
200 = 𝑉 × 1
𝑉 = 200 𝑐𝑚3
But 𝐿3 = 𝑉
𝐿3 = 200
𝐿 = 5.85 𝑐𝑚

42. HYDROMETER
• This is an instrument used to measure relative density or consist off a uniform glass tube
containing lead shots. In another form of the instrument a uniform stem is mounted on a
large bulb that has lead shots in it as shown in the diagram below

43. HYDROMETER
Calculations
When height increases the 𝜌 reduces
𝑀 = 𝜌𝐴ℎ
𝑀
𝐴ℎ
= 𝜌 . 𝜌 =
1
ℎ
calibration is non linear
Calculation in stem glass

44. WEIGHTLESSNESS – MOTION IN A LIFT
When a lift accelerates downwards occupants feel lighter, however they feel heavier
when it accelerates upwards. These changes can be explained by considering the
resultant force, which acts on the occupants as a combination of two forces that are
acting. These are:
i. Force of gravity on weight
ii. The force needed to accelerate or decelerate the lift

45. WEIGHTLESSNESS – MOTION IN A LIFT
If the lift accelerates downwards with an acceleration of 𝑎ms−2
then 𝑾 = 𝒎𝒈 − 𝒎𝒂. If
the lift accelerates upwards with acceleration of 𝑎ms−2
then 𝑾 = 𝒎𝒈 + 𝒎𝒂 which
implies that the weight appears to be increased and occupants appear to be decreased.
As the lift increases its downwards acceleration, the apparent weight will be less and
less until eventually becomes zero. If the acceleration of the lift increases beyond 10
𝑚/𝑠2 then the person inside will fly. Spacemen experience weightlessness when the
acceleration of their spacecraft is greater than or equal to acceleration due to
gravity.

46. QUESTION
• Calculate the force with which the feet of a passenger passes downwards on the floor of an
elevator accelerating upwards of 4 × 10−3
𝑚𝑠−2
if the passenger’s weight is 60 𝑁.
SOLUTION
Weight of passenger = 60 N
But weight of passenger = 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐦𝐚𝐧 × 𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧 𝐝𝐮𝐞 𝐭𝐨 𝐠𝐫𝐚𝐯𝐢𝐭𝐲
Mass of passenger =
𝐰𝐞𝐢𝐠𝐡𝐭 𝐨𝐟 𝐩𝐚𝐬𝐬𝐞𝐧𝐠𝐞𝐫
𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧 𝐝𝐮𝐞 𝐭𝐨 𝐠𝐫𝐚𝐯𝐢𝐭𝐲
I.E 𝒎 =
𝑾
𝒈
=
𝟔𝟎
𝟏𝟎
= 𝟔 𝒌𝒈.
Let 𝑊𝑎 be the apparent weight of the passenger
⇒ 𝑊𝑎 = 𝑚𝑔 + 𝑚𝑎
⇒ 𝑊𝑎 = 6 × 10 + 6 × (4 × 10−3
)
⇒ 𝑊𝑎 = 𝟔𝟎. 𝟎𝟐𝟒 𝑁

47. QUESTION
• A man of mass 70 kg is standing in a lift. What force does the floor of the lift exert on the man if
the lift is
i. moving with a uniform velocity?
ii. accelerating at 3 𝑚𝑠−2 upwards?
iii. Accelerating at 3 𝑚𝑠−2
downwards? (Take 𝑔 = 10 𝑚𝑠−2
)

48. SOLUTION
R is the normal reaction from the floor on the man
i)
Since the lift is moving with a uniform velocity, the resultant force is zero: 𝑅 = 𝑚𝑔 = 70 × 10 =
700 𝑁
R
mg
a
Since the lift is accelerating upwards
Equation of motion: 𝑅 − 𝑚𝑔 = 𝑚𝑎
⇒ 𝑅 = 𝑚𝑔 + 𝑚𝑎 = 𝑚 𝑔 + 𝑎 = 70 10 + 3 = 70 × 13 = 𝟗𝟏𝟎 𝑵

49. ii)
Since the lift is accelerating downwards
Equation of motion: 𝑚𝑔 − 𝑅 = 𝑚𝑎
⇒ 𝑅 = 𝑚𝑔 − 𝑚𝑎 = 𝑚 𝑔 − 𝑎 = 70 10 − 3 = 70 × 7 = 𝟒𝟗𝟎 𝑵

50. CONNECTED BODIES
Two particles connected by a light inextensible string passing over a fixed light smooth
frictionless pulley are called connected bodies. The tension in the string is the same
throughout its length so the body is acted upon by the same tension. Problems concerned
with connected bodies usually involve finding the acceleration of the system and the
tension in the string.

51. BODIES ON HORIZONTAL SURFACES
𝒇 𝒙
𝑓𝑥 = 𝜇𝑅 = 𝜇𝑚1 𝑔
For 𝒎 𝟏
𝑚1 𝑎 = 𝑇 − 𝑓𝑥 or 𝑚1 𝑎 = 𝑇 − 𝜇𝑚1 𝑔
For 𝒎 𝟐
𝑚2 𝑎 = 𝑚2 𝑔 − 𝑇
Eqn (1) + eqn(2)
𝑚1 𝑎 + 𝑚2 𝑎 = 𝑚2 𝑔 − 𝜇𝑚1 𝑔
𝑎 𝑚1 + 𝑚2 = 𝑚2 𝑔 − 𝜇𝑚1 𝑔
𝒂 =
𝒎 𝟐 𝒈
𝒎 𝟏 + 𝒎 𝟐
−
𝝁𝒎 𝟏 𝒈
𝒎 𝟏 + 𝒎 𝟐

52. BODIES ON INCLINED PLANES
𝑓𝑥 = 𝜇𝑅 = 𝜇𝑚2 𝑔 𝑐𝑜𝑠𝜃
For 𝒎 𝟏
𝑚1 𝑎 = 𝑚1 𝑔 − 𝑇 − (1)
For 𝒎 𝟐
𝑚2 𝑎 = 𝑇 − [𝑓𝑥 + 𝑚2 𝑔 sin 𝜃 ]
Eqn (1) + eqn(2)
𝑚1 𝑎 + 𝑚2 𝑎 = 𝑚1 𝑔 − (𝑓𝑥 + 𝑚2 𝑔𝑠𝑖𝑛 𝜃)
𝑎 𝑚1 + 𝑚2 = 𝑚1 𝑔 − (𝑓𝑥 + 𝑚2 𝑔𝑠𝑖𝑛 𝜃)
𝒂 =
𝒎 𝟏 𝒈
𝒎 𝟏 + 𝒎 𝟐
−
(𝒇 𝒙 + 𝒎 𝟐 𝒈𝒔𝒊𝒏 𝜽)
𝒎 𝟏 + 𝒎 𝟐

53. BODIES ON INCLINED PLANES
𝑓𝑥 = 𝜇𝑅 = 0.2 × 2 × 10 cos 30 = 3.5
For 𝒎 𝟏
𝑚1 𝑎 = 𝑚1 𝑔 − 𝑇
𝟏𝟎𝒂 = 𝟏𝟎𝟎 – 𝑻 __ (𝟏)
For 𝒎 𝟐
𝑚2 𝑎 = 𝑇 − [𝑓𝑥 + 𝑚2 𝑔 sin 𝜃 ]
2𝒂 = 𝑻 – (𝟑. 𝟓 + 𝟐 × 𝟏𝟎𝒔𝒊𝒏 𝟑𝟎
2𝒂 = 𝑻 − 𝟏𝟑. 𝟓 __ (𝟐)
Eqn (1) + eqn(2)
12𝑎 = 100 − 13.5
𝒂 = 𝟕. 𝟐𝟏𝒎/𝒔 𝟐
Substituting a = 7.21 into eqn 1
𝟏𝟎 𝟕. 𝟐𝟏 = 𝟏𝟎𝟎 − 𝑻
𝑻 = 𝟐𝟕. 𝟗
= 10 kg
𝟑𝟎°
= 2 kg
𝝁 = 𝟎. 𝟐

54. PULLEY SYSTEM
Because the direction 𝑚1 𝑔 is greater than the 𝑚2
𝑚1 > 𝑚2
For 𝒎 𝟏
𝑚1 𝑎 = 𝑚1 𝑔 − 𝑇 − (1)
For 𝒎 𝟐
𝑚2 𝑎 = 𝑇 − 𝑚2 𝑔 − 2
Eqn (1) + eqn(2)
𝑚1 𝑎 + 𝑚2 𝑎 = 𝑚1 𝑔 − 𝑚2 𝑔
𝑎 𝑚1 + 𝑚2 = 𝑚1 𝑔 − 𝑚2 𝑔
𝒂 =
𝒎 𝟏 𝒈
𝒎 𝟏 + 𝒎 𝟐
−
𝒎 𝟐 𝒈
𝒎 𝟏 + 𝒎 𝟐

55. A thread is passed over a pulley. 10 kg mass and 8 kg mass are suspended at the ends of
the ropes. Draw the arrangement and indicate the body force diagram on the masses.
Evaluate
1. The acceleration of either masses
2. The tension in the tie
Solution
For 𝒎 𝟏
10𝑎 = 100 − 𝑇 − (1)
For 𝒎 𝟐
8𝑎 = 𝑇 − 80 − 2
Eqn (1) + eqn(2)
18𝑎 = 20
𝑎 =
20
18
=
10
9
= 1.11 𝑚/𝑠2
Substitute a into eqn 1
𝑻 = 𝟖𝟖. 𝟗 𝑵