The Wayback Machine - https://web.archive.org/web/20230310185200/https://www.geeksforgeeks.org/problems-on-friction-formula/
Skip to content
Related Articles
Open in App
Not now

Related Articles

  • Problems on Friction Formula
  • Plant Kingdom
  • Algae – Definition, Characteristics, Types and Examples
  • Chlorophyceae
  • Phaeophyceae – Overview, Characteristics, Importance, Examples
  • Rhodophyceae
  • Bryophytes | Class 11 Biology
  • Liverworts
  • Mosses
  • Pteridophytes
  • Gymnosperms – Definition, Characteristics, Uses and Examples
  • Angiosperms
  • Difference between Angiosperms and Gymnosperms
  • Problems on Friction Formula
  • Biomolecules – Definition, Structure, Classification, Examples
  • What are Metabolites? – Primary and Secondary
  • Biomacromolecules – Definition, Types, Functions, Significance
  • Proteins – Definition, Structure, Significance, Examples
  • Polysaccharides – Definition, Types, Features, Examples
  • Nucleic Acid – Definition, Function, Structure, and Types
  • Protein Structure – Primary, Secondary, Tertiary, Quaternary
  • Metabolic Basis For Living | CBSE Class 11 Biology Chapter 9
  • Enzymes – Definition, Structure, Classification, Examples
  • How Do Enzymes Bring About Such High Rates Of Chemical Conversions?
  • Nature of Enzyme Action
  • Mechanism of Enzymes Action
  • Factors Affecting Enzyme Activity
  • Cofactors – Definition, Structure, Types, Examples
  • Problems on Friction Formula
  • Respiration In Plants
  • Do Plants Breathe?
  • Glycolysis
  • Fermentation
  • What is Aerobic Respiration? – Definition, Diagram, Process
  • Tricarboxylic Acid Cycle – Overview, Stages, Roles, Significance
  • Electron Transport System (ETS) And Oxidative Phosphorylation
  • Amphibolic Pathway
  • Respiratory Quotient
  • Problems on Friction Formula
  • Human Respiratory System
  • Inspiration and Expiration
  • Lung Volumes And Capacities
  • Transport of Carbon Dioxide in the Blood
  • Regulation Of Respiration
  • Respiratory System Disorders – Definition, Causes, Types, Symptoms
  • Problems on Friction Formula
  • Body Fluids and Circulation
  • Blood Groups – ABO Blood Group and Rh Group System
  • Blood Coagulation
  • Lymphatic System
  • Circulatory Pathways – Anatomy and Functions
  • Cardiac Cycle | Class 11 Biology
  • Electrocardiogram (ECG)
  • Double Circulation
  • Regulation of Cardiac Activity
  • Disorders of the Circulatory System
  • Problems on Friction Formula
  • Reflex Actions – Definition, Process, Examples, FAQs
  • Anatomy and Physiology of Human Eye
  • Anatomy and Physiology of Human Ear
  • Problems on Friction Formula
  • Endocrine Gland
  • Hypothalamus – Function, Hormones and Disorder
  • Pituitary Gland
  • Pineal Gland – Definition, Structure, Importance, Functions
  • Thyroid Gland – Anatomy, Function and Clinical Aspects
  • Parathyroid Gland – Functions and Disorders
  • What is Thymus Gland?
  • Adrenal Gland
  • Pancreas
  • Testes – Anatomy and Functions
  • Ovary – Female Reproductive System
  • CBSE Class 11 | Computer Science – C++ Syllabus
  • Problems on Friction Formula
  • Respiration In Plants
  • Do Plants Breathe?
  • Glycolysis
  • Fermentation
  • What is Aerobic Respiration? – Definition, Diagram, Process
  • Tricarboxylic Acid Cycle – Overview, Stages, Roles, Significance
  • Electron Transport System (ETS) And Oxidative Phosphorylation
  • Amphibolic Pathway
  • Respiratory Quotient
  • Problems on Friction Formula
  • Human Respiratory System
  • Inspiration and Expiration
  • Lung Volumes And Capacities
  • Transport of Carbon Dioxide in the Blood
  • Regulation Of Respiration
  • Respiratory System Disorders – Definition, Causes, Types, Symptoms
  • Problems on Friction Formula
  • Body Fluids and Circulation
  • Blood Groups – ABO Blood Group and Rh Group System
  • Blood Coagulation
  • Lymphatic System
  • Circulatory Pathways – Anatomy and Functions
  • Cardiac Cycle | Class 11 Biology
  • Electrocardiogram (ECG)
  • Double Circulation
  • Regulation of Cardiac Activity
  • Disorders of the Circulatory System
  • Problems on Friction Formula
  • Reflex Actions – Definition, Process, Examples, FAQs
  • Anatomy and Physiology of Human Eye
  • Anatomy and Physiology of Human Ear
  • Problems on Friction Formula
  • Endocrine Gland
  • Hypothalamus – Function, Hormones and Disorder
  • Pituitary Gland
  • Pineal Gland – Definition, Structure, Importance, Functions
  • Thyroid Gland – Anatomy, Function and Clinical Aspects
  • Parathyroid Gland – Functions and Disorders
  • What is Thymus Gland?
  • Adrenal Gland
  • Pancreas
  • Testes – Anatomy and Functions
  • Ovary – Female Reproductive System
  • CBSE Class 11 | Computer Science – C++ Syllabus
  • Problems on Friction Formula

    Improve Article
    Save Article
    • Difficulty Level : Expert
    • Last Updated : 14 Jul, 2021
    Improve Article
    Save Article

    Friction is a force that is present almost everywhere in daily life. This force allows us to walk, hold objects, etc. In short, it is essential for almost all physical actions in real life. When a ball is thrown on the floor with some velocity, it keeps moving for some time and then comes to a halt. In an ideal scenario, the ball should keep moving as there is no external force acting on the ball. But it comes to rest, which means that there is a force present that resists the motion. This force is called friction. Frictional force by nature opposes all sorts of motion between objects. 

    Friction

    Frictional forces oppose the motion. While it usually generates heat and wears and tear objects, it is often helpful. It gives the necessary traction needed for walking and running without slipping. In automobiles, it is less desired as it accounts for 20% of energy loss. Friction depends upon the normal force acting on the surface. Frictional forces can be divided into two parts: 

    1. Static Friction: This is a type of friction that prevents objects from sliding on a surface. The direction stops the objects from sliding against one another and is parallel to the surface.
    2. Kinetic Friction: This type of friction acts when an object slides on the surface. The direction is the same as static friction.

    Laws of Friction: 

    1. Friction of the moving object is proportional and perpendicular to the normal force applied on it.
    2. The friction experienced by the object depends on the nature of the surface it is in contact with.
    3. Friction is independent of the area of contact as long as there is an area of contact.
    4. Kinetic friction is independent of velocity.
    5. The coefficient of static friction is greater than the coefficient of kinetic friction.

    Equations of Friction

    When a force is applied to the object, let’s say FN is the force acting normally on the surface. The value of the static friction is given by Fs and \mu_s   is the value of the coefficient of static friction. Then the equation governing the static force of friction is given by, 

    F_s \leq \mu_sF_N

    In the case where the object is moving, the kinetic force of the friction will be given by, 

    F_k = \mu_kF_N

    The coefficient of friction is given by, 

    \frac{F}{N} = \mu

    Here, F is the force of friction and N is the value of the normal force. 

    Let’s look at some problems related to these concepts. 

    Sample Problems

    Question 1: Force of 60N is exerted on the box kept on the floor with the coefficient of static friction of 0.4. Find the friction force.  

    Answer: 

    Force of static friction is given by, 

    F_s = \mu_sF_N

    Here, \mu_s is the coefficient of static friction and N is the normal force. 

     \mu_s = 0.4 and N = 60N 

    Plugging the values in the equation, 

    F_s = \mu_sF_N \\ = F_s = (0.4)(60) \\ = F_s = 24 \text{ N}

    Question 2: Force of 100N is exerted on the box kept on the floor with the coefficient of static friction of 0.2. Find the friction force

    Answer: 

    Force of static friction is given by, 

    F_s = \mu_sF_N

    Here, \mu_s is the coefficient of static friction and N is the normal force. 

     \mu_s = 0.2 and N = 100N 

    Plugging the values in the equation, 

    F_s = \mu_sF_N \\ = F_s = (0.2)(100) \\ = F_s = 20 \text{ N}

    Question 3: Force of 100N is exerted at an angle of 30° on the box kept on the floor with the coefficient of static friction of 0.2. Find the friction force

    Answer: 

    Force of static friction is given by, 

    F_s = \mu_sF_N

    Here, \mu_s   is the coefficient of static friction and N is the normal force. 

     \mu_s  = 0.2, \theta = 30° and F = 100N

    N = Fsin(\theta

    ⇒N = (100)sin(30°) 

    ⇒N = 50

    Plugging the values in the equation, 

    F_s = \mu_sF_N \\ = F_s = (0.2)(50) \\ = F_s = 10 \text{ N}

    Question 4: Force of 200N is exerted at an angle of 30° on the box kept on the floor with the coefficient of static friction of 0.2. Find the friction force

    Answer: 

    Force of static friction is given by, 

    F_s = \mu_sF_N

    Here, \mu_s is the coefficient of static friction and N is the normal force. 

     \mu_s  = 0.2, \theta = 30° and F = 200N

    N = Fsin(\theta

    ⇒N = (200)sin(30°) 

    ⇒N = 100

    Plugging the values in the equation, 

    F_s = \mu_sF_N \\ = F_s = (0.2)(100) \\ = F_s = 20 \text{ N}

    Question 5: Force of 60N is exerted horizontally on the box of 40Kg kept on the floor with the coefficient of static friction of 0.2. Find the friction force. 

    Answer: 

    Force of static friction is given by, 

    F_s = \mu_sF_N

    Here, \mu_s is the coefficient of static friction and N is the normal force. 

     \mu_s = 0.2 and N = mg

    N = mg 

    ⇒N = (40)(10)

    ⇒N = 400N 

    Static friction will be, 

    F_s = \mu_sF_N \\ F_s = (0.2)(400) \\ F_s = 80 \text{ N}

    Now, the applied force is 60N and the maximum static friction is 80N. Since the applied force is less than the value of static friction, the box will not move and the frictional force will be equal to the applied force. 

    Question 6: Force of 50N is exerted horizontally on the box of 50Kg kept on the floor with the coefficient of static friction of 0.3. Find the friction force. 

    Answer: 

    Force of static friction is given by, 

    F_s = \mu_sF_N

    Here, \mu_s is the coefficient of static friction and N is the normal force. 

     \mu_s = 0.3 and N = mg

    N = mg 

    ⇒N = (50)(10)

    ⇒N = 500N 

    Static friction will be, 

    F_s = \mu_sF_N \\ F_s = (0.3)(500) \\ F_s = 150 \text{ N}

    Now, the applied force is 100N and the maximum static friction is 150N. Since the applied force is less than the value of static friction, the box will not move and the frictional force will be equal to the applied force. 

    Question 7:A horizontal force of 40N is exerted on a box of 2Kg kept on the floor with the coefficient of static friction of 0.4 and a coefficient of kinetic friction is 0.2. Find the force of friction. 

    Answer: 

    Force of static friction is given by, 

    F_s = \mu_sF_N

    Here, \mu_s is the coefficient of static friction and N is the normal force. 

     \mu_s = 0.4 and N = mg 

    N = mg 

    ⇒N = (2)(10) 

    ⇒N = 20N 

    Plugging the values in the equation, 

    F_s = \mu_sF_N \\ = F_s = (0.4)(20) \\ = F_s = 8 \text{ N}

    Now, since the maximum static friction is less than the force applied. The box will start moving and then the kinetic friction will act on the body.

    F_k = \mu_kF_N\\ = F_k = (0.2)(mg) \\ = F_k = (0.2)(2 \times 10) \\ = F_k = 4 \text{ N}

    Question 8:A horizontal force of 50N is exerted on a box of 3Kg kept on the floor with the coefficient of static friction of 0.4 and a coefficient of kinetic friction is 0.2. Find the force of friction. 

    Answer: 

    Force of static friction is given by, 

    F_s = \mu_sF_N

    Here, \mu_s is the coefficient of static friction and N is the normal force. 

     \mu_s = 0.4 and N = mg 

    N = mg 

    ⇒N = (3)(10) 

    ⇒N = 30N 

    Plugging the values in the equation, 

    F_s = \mu_sF_N \\ = F_s = (0.4)(30) \\ = F_s = 12 \text{ N}

    Now, since the maximum static friction is less than the force applied. The box will start moving and then the kinetic friction will act on the body.

    F_k = \mu_kF_N\\ = F_k = (0.2)(mg) \\ = F_k = (0.2)(3 \times 10) \\ = F_k = 6 \text{ N}


    My Personal Notes arrow_drop_up
    Related Articles

    Start Your Coding Journey Now!