Video Lesson

Lesson objective

Dear learners,

At the end of the lesson you will be able to

• Describe the general characteristics of friction
• List the various types of friction
• Calculate the magnitude of static and kinetic friction, and use these in problems involving Newton’s laws of motion
• Solve more complex acceleration problems involving Newton’s laws of motion.
• Use free-body diagrams to solve problems on Newton’s Laws of motion.

Brainstorming question

• What is the role of friction in our daily life?
• Can you get rid of friction?
• Mention pros and cons of frictional force?

Key Terms and Concepts

Definition of friction

• Friction is the force that resists the relative motion of material elements, solid surfaces, fluid layers from rolling or sliding against each other. 
• Friction basically arises from the action of irregularities of the harder surface plowing across the softer one.
• Friction is a type of force that We experience in our everyday activity.
• Friction, in simple terms is a force that opposes motion.
• It occurs when two surfaces are in contact with each other and when one surface slides or attempts to slide over the other.
• Friction occurs due to the roughness of the surfaces in contact. Even if the surface seem very smooth,
• it is not so at micro or Nano scale,
• Friction helps us to move while we walk by preventing us from slipping.
• Without friction between the tires and the road we couldn’t drive or turn the car.
• Without the frictional force exerted by the air on a body moving through it (air drag),
• parachutes do not work.
• Without friction nails would pull out,
• light bulbs would unscrew effortlessly.

Types of friction

• There are two types of friction: Static friction and Kinetic friction.
• Static friction is the friction between two surfaces when there is no movement. Suppose you pull a block slightly along a table top, Figure 4.19a. The block will not move with such a small force that you apply. What do you think is the reason for this? The force that keeps the block form sliding is the force of static friction denoted by (F_S ) and is directed opposite to the applied force.
• Kinetic friction: Kinetic friction is the friction between two surfaces when one of them is sliding over the other. For example, when the block that you push begins to slide over the table, Figure 4.19b, there is force of kinetic friction denoted by (F_K) between the bottom surface of the block and the table top
• Frictional force is directly related to the normal force and its value depends upon the property of the surfaces in contact.
• This property of material that resists motion µ is called the coefficient of friction ($\mu$ ), which is defined as the ratio between the friction force and the normal force. Mathematically,
• $Coefficient of friction=\frac{Frictional force}{Normal force}$
• or symbolically $\mu=\frac{F_{f}}{F_{n}}$
• Coefficient of friction is a dimensionless number.
• We write F_S = µ_sF_n , for static frictional force, where µ_s is coefficient of static friction
• F_K = µ_KF_n for kinetic frictional force, where µ_K is coefficient of kinetic friction

Figure 4.1 Static friction and kinetic friction

The maximum value of static

• The maximum value of static friction is known as limiting friction and it is the frictional force between the surfaces when the body just begins to slide.
• Static friction is less than or equal to the product of coefficient of static friction (µs) and the normal force (F_N).
• $F_{k}\le \mu_{s}F_{N}$
• The maximum value being F_S = µ_sF_n
• Once the static friction gives way to kinetic friction, i.e., the block is in motion, it is easier to keep it in motion than it was to get it started, indicating that the kinetic frictional force is less than the static frictional force.
• Kinetic friction is less than static friction.
• $F_{k}\lt F_{s}$ and $\mu_{k}\lt \mu_{s}$

Note that

• Unlike the kinetic friction force, the static friction force takes on any value between zero and its maximum value of µ_s F_N , depending on the magnitude of the applied
• force. It could take any value between zero to its limiting value.
• 0≤ F_s≤µ_s F_N

Plotting frictional force versus applied force

Examples 3.1

An 8 kg block is placed on a horizontal surface. The coefficient of static friction and that of kinetic friction between the block and surface are 0.4, and 0.35, respectively.

A. What is the horizontal force just enough to start moving the block?

B. What horizontal force must be applied on the block to keep it uniformly accelerating at 4 m/s²?

Given:

• m = 8 kg, µ = 0.4, µ = 0.3

Solution

A. The block is in a limiting equilibrium, so that the friction force will be , F_s= µ_s F_N

• (since the block is on a level surface F_N=mg
• Therefore, F = µ mg = (0.4) (8 kg) (9.8 m/s ) = 31.4 N

B. As the block accelerates the net force that accelerates the block is the vector sum of the applied force and the kinetic frictional force.

• F_net= F_app− F_k
• F_net = ma = (8 kg )(4 m / s ) = 32 N2 and
• F_k= µ_kmg = (0.3) (8 kg) (9.8 m/s ) = 23.5 N
• Thus, F_app = F_net + F_k =32 N + 23.5 N = 55.5 N

Example 3.2

A 6 kg block is placed on a ramp that makes an angle of 30 above the horizontal Figure 4.23a. If the block remains at rest,
A) what is the static friction that supports the block from sliding down the ramp?
B) what is the coefficient of static friction between the block and the ramp?

• As shown in Figure 4.23b, the force that tends to move the block down the ramp is mg sinθ and the one that prevented sliding is static friction F_s . The normal force between the ramp and the block is F_N = mg cosθ.
• A) Under the condition where the block is at rest we write,
• F _net = 0
• mg sinθ-F_s = 0
• F_s = mg sinθ = (6 kg)(9.8 m/s )(sin 30 ) = 29.4 N
• Fs = µs FN = µs mg cosθ
• µ_s=$\frac{F_{s}}{mgcos\Theta}$
• µ_s=$\frac{29.4N}{\left( 6kg \right)\left( 9.8m/s^{2} \right)\left( cos30^{0} \right)}$=0.58
• See that mg sin θ− F_s = 0,
• we have mg sin θ−µ_s mg cos θ =0
• $\mu_{s}=\frac{sin\Theta}{cos\Theta}=tan\Theta$=
• tan30⁰ =0.58

Figure 4.3 A block on a ramp

Example 3.3

Two blocks of identical materials are connected by a light string on a level surface (Figure 4.25). Assuming no friction between the blocks and the level surface, find the acceleration of the masses and the tension in the connecting string when the string attached to m is pulled to the right by a 36 N force. Use m = 4 kg and m = 8 k

Solution:

Given:

$m_1 = 4 kg, m_2 = 8 kg, F = 36 N$
The blocks move together to the right under the action of force F. Writing Newton’s second law for each block we have
For $m_1, F-T =m_1 a$
For $m_2, T = m_2 a$
Adding the above equations, we get
$F = m_1 a + m_2 a = (m_1 + m_2 )a$
$a = \frac {F}{(m_1 + m_2 )}$
Substituting the given values, we obtain $a = 3 m/s^2$. Using a $3 m/s^2$ in $T = m_2a$ we get T = 24 N

Tension Forces

• The tension force acts along the direction of the string and exerts a force both on the object and on the person exerting force on the cable, as illustrated in Figure 4.28.
• When an object attached to a string (or a cable) is pulled by means of a pulling force exerted on the string, the force on the string is called a tension force.
• If the mass of the given cable is neglected, the same tension acts on both the object and the person

Vertical Tension force on a static object

The string attached to an object is pulled upward under such tension that it accelerates the object upward (Figure 4.30). In this case the tension supports the weight and also accelerates the object upward. Applying Newton’s second law along the y axis we write

Figure 4.5 An object hangs by a string in an elevator accelerating upward