Lecture-01

BOOK CHAPTER 5

(Force and Motion-1)

Force:
❑ A force is a push or a pull.
❑ A force is an interaction between two objects.
❑ A force is a vector quantity, with magnitude and direction.
Units of force: SI unit: Newton (N); CGS unit: dyne; British unit: pound (lb)
If two or more forces act on a body, we find the net force (or resultant force) by
adding them as vectors.
𝐹Ԧ = 𝐹𝑥 𝑖Ƹ + 𝐹𝑦 𝑗Ƹ

The force , which acts at an angle from the x-axis, may be replaced by its rectangular component
vectors 𝐹Ԧ𝑥 and 𝐹Ԧ𝑦 . 𝐻𝑒𝑟𝑒 𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃 𝑎𝑛𝑑 𝐹𝑦 = 𝐹𝑠𝑖𝑛𝜃 .
 Some Particular Forces:
The Gravitational Force:
A gravitational force on a body is a pull by another body. In most situations, the other body is
Earth or some other astronomical body. For Earth, the force is directed down toward the ground,
which is assumed to be an inertial frame.
With that assumption, the magnitude of 𝐹Ԧ𝑔 is
𝐹𝑔 = 𝑚𝑔
where m is the body’s mass and g is the magnitude of the free-fall acceleration.
Normal Force:
A normal force 𝑭𝑵 is the force on a
body from a surface against which
the body presses. The normal force is
always perpendicular to the surface.

Figure (a) A block resting on a table experiences a normal force

perpendicular to the tabletop. (b) The free-body diagram for the block.
Weight:
The weight W of a body is equal to the magnitude 𝐹𝑔 of the gravitational force on the
body. That is,
𝑊 = 𝐹𝑔 = 𝑚𝑔

Frictional force:
A frictional force is the force on a body
when the body slides or attempts to slide
along a surface. The force is always
parallel to the surface and directed so as
to oppose the sliding. On a frictionless
surface, the frictional force is negligible.
Tension:
When a cord (or a rope, cable, or other such object) is attached to a body and pulled taut, the cord pulls
on the body with a force directed away from the point of attachment to the body and along the cord (as
shown in the adjacent figure). The force is often called a tension force.
For a massless cord (a cord with negligible mass), the pulls at both ends of the cord have the same
magnitude T, even if the cord runs around a massless, frictionless pulley (a pulley with negligible mass
and negligible friction on its axle to oppose its rotation).
Newtonian Mechanics:
The relation between a force and the acceleration it causes was first understood by Isaac
Newton (1642 –1727) .The study of that relation, as Newton presented it, is called
Newtonian mechanics. We shall focus on its three primary laws of motion.
Newton’s First Law:
If there is no net force on a body, the body remains at rest if it is initially at rest or moves in a
straight line at constant speed if it is in motion.
OR
If no net force acts on a body (𝑭𝒏𝒆𝒕 = 𝟎), the body’s velocity cannot change; that is,
the body cannot accelerate.
Newton’s Second Law:
The net force(𝐹Ԧ𝑛𝑒𝑡 ) on a body is equal to the
product of the body’s mass (m) and its acceleration
(𝑎).
Ԧ For a specific case along x-axis,
In vector equation form,
𝐹𝑛𝑒𝑡,𝑥 = 𝑚𝑎𝑥
𝐹Ԧ𝑛𝑒𝑡 = 𝑚𝑎Ԧ 𝐹1 − 𝐹2 = 𝑚𝑎𝑥
 Units in Newton's Second Law

Newton’s Third Law:

When two bodies interact, the forces on the
bodies from each other are always equal in
magnitude and opposite in direction.
For the book and crate, we can write this
law as the vector relation

𝐹Ԧ𝐵𝐶 = −𝐹Ԧ𝐶𝐵 The force on B due to C has the

same magnitude as the force on C
(equal magnitudes and opposite directions) due to B.
Problem 5.45: An elevator cab that weighs 27.8 kN moves upward. What is the
tension in the cable if the cab’s speed is (a) increasing at a rate of 1.22 𝑚𝑠 −2 and (b)
decreasing at a rate of 1.22 𝑚𝑠 −2 ?

Answer: The mass of the elevator, 𝑚 = (27800/9.80)𝑘𝑔 = 2837 𝑘𝑔

Acceleration, 𝑎 = +1.22𝑚𝑠 −2 (with +y upward)

Newton’s second law leads to

𝑇 − 𝑚𝑔 = 𝑚𝑎
𝑇 = 3.13 ∗ 104 𝑁
(b) The term “deceleration” means the acceleration vector is in the direction opposite
to the velocity vector (which the problem tells us is upward). Thus (with +y upward)
the
acceleration is now 𝑎 = – 1.22 𝑚𝑠 −2
So that the tension is
𝑇 = 𝑚 (𝑔 + 𝑎) = 2.43 ∗ 104 𝑁 .
Sample Problem 5.03: Figure shows a block S (the sliding block) with mass M 3.3 kg. The block is free to
move along a horizontal frictionless surface and connected, by a cord that wraps over a frictionless pulley, to a
second block H (the hanging block), with mass m 2.1 kg. The cord and pulley have negligible masses
compared to the blocks (they are “massless”). The hanging block H falls as the sliding block S accelerates to
the right. Find (a) the acceleration of block S, (b) the acceleration of block H, and (c) the tension in the cord.
Answer: For S block, there is only one force component, which is T. Therefore,
𝑇 = 𝑀𝑎………..(1)
For the hanging block H, the acceleration is downward and the tension of the cord acts upward.
According to Newton’s second law,
𝑚𝑔 − 𝑇 = 𝑚𝑎………(2)
Substituting the value of T from (1) in eqn (2)
𝑚𝑔 − 𝑀𝑎 = 𝑚𝑎

𝑚𝑔
𝑎=
𝑚+𝑀

2.1 ∗ 9.8
𝑎= 𝑚𝑠 −2
2.1 + 3.3

𝑎 = 3.8 𝑚𝑠 −2
Substituting the value of a in eqn (1)
𝑇 = (3.3 ∗ 3.8) 𝑁 = 12.6𝑁
Problem: A 10 kg mass is attached to the ceiling with two
strings (shown below). If the system is in static equilibrium,
what is the tension in each of the strings?

Answer: Along the X axis,

𝑇1 𝑐𝑜𝑠45° − 𝑇2 𝑐𝑜𝑠45° = 0
𝑇1 = 𝑇2 T1sin45o
T2sin45o

Along the Y axis,

𝑇1 𝑠𝑖𝑛45° + 𝑇2 𝑠𝑖𝑛45° − 𝑚𝑔 = 0 T1cos45o TT2cos45
2cos45
oo

𝑇1 = 𝑇2 = 70.7N

mg
Homework
Problem 5.51: Figure shows two blocks connected by a cord (of negligible mass) that passes over a
frictionless pulley (also of negligible mass). The arrangement is known as Atwood’s machine. One
block has mass m1 =1.30 kg; the other has mass m2 = 2.80 kg. What are (a) the magnitude of the
blocks’ acceleration and (b) the tension in the cord?
Hints:
𝑇 − 𝑚1𝑔 = 𝑚1𝑎
𝑀2𝑔 − 𝑇 = 𝑚2𝑎
Ans: (a) 3.6m/s2 (b) 17.4 N
Problem 5.61: A hot-air balloon of mass M is descending vertically with downward acceleration of
magnitude a. How much mass must be thrown out to give the balloon an upward acceleration of magnitude
a? Assume that the upward force from the air (the lift) does not change because of the decrease in mass.
Hints: Upward force = Fa
Before the mass is thrown out
𝑀𝑔 − 𝐹𝑎 = 𝑀𝑎
After the mass is thrown out, the mass is M – m.
𝐹𝑎 – (𝑀 − 𝑚)𝑔 = (𝑀 − 𝑚)𝑎
2𝑀𝑎
Ans :
𝑔+𝑎