Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this document? Why not share!

No Downloads

Total views

4,555

On SlideShare

0

From Embeds

0

Number of Embeds

8

Shares

0

Downloads

0

Comments

0

Likes

2

No embeds

No notes for slide

- 1. FRICTION J3010/4/1 UNIT 4FRICTION OBJECTIVESGeneral Objective : To understand the concept of friction.Specific Objectives : At the end of this unit you should be able to : recognize the difference between Coefficient of friction and Angle of friction. illustrate these forces involved in motion on the inclined plane and horizontal plane by diagram. use the formulae to solve of problem on friction. calculate the answers using the concept of friction.
- 2. FRICTION J3010/4/2 INPUT 4.0 INTRODUCTION When two bodies are in contact with one another, due to gravitational or external forces, a force called the frictional force is exerted between the contacting surfaces which tends to prevent movement between them. This tendency to prevent sliding is called friction. When a person is walking there is a tendency for the feet to slip backwards. Friction between the feet and the ground, which acts in the forward direction prevents the feet from slipping Coefficient of Friction It is the ratio of the limiting friction to the normal reaction, between the two bodies and generally denoted µ . Mathematically: F = µR N where µ is the term coefficient of friction and it is constant. Fig. 4.1
- 3. FRICTION J3010/4/3 4.1 MOTION UP THE PLANE: PULL P PARALLEL TO PLANE The friction force F opposes the motion of the body up the plane. Fig. 4.2 The angle between W and RN is θ , the angle of the plane. Resolving perpendicular to the plane: R N = W cos θ Resolving parallel to the plane: P = F + W sin θ and for limiting friction: F = µRN Example 4.1 Find the pull required to move a load of 30 kg up a rough plane, the pull being applied parallel to the plane. The inclination of the plane at 30˚ and pull of 6 kg to a similar smooth inclined plane, would keep the same load in equilibrium. The coefficient of friction is 0.3.
- 4. FRICTION J3010/4/4 Solution 4.1 Fig. 4.3 Given: W = 30 kg μ = 0.3 Pull, P = 6 kg α = 30˚ Force of friction, F = 0 (smooth surface) Let angle of inclination = θ Resolving the force along the inclined plane, cos 30˚ = 30 sin θ 6 cos 30 0 sin θ = 30 6x0.866 = 30 = 0.1732 θ = 9˚ 58’ Resolving the forces along the inclined plane, P = F + 30 sin 9˚ 58’ = 0.3 R + 30 sin 9˚ 58’ ----- (i) Resolving the forces at right angle to the plane, P = 30 cos 9˚ 58’ ----- (ii) Substituting the value of R in equation (i) P = 0.3 x 30 cos 9˚ 58’ + 30 sin 9˚ 58’ = 0.3 x 30 (0.9849) + 30 (0.1732) = 14.06 kg
- 5. FRICTION J3010/4/5 4.2 MOTION DOWN THE PLANE: PULL P PARALLEL TO PLANE Fiq. 4.4 When the body is pulled down the plane by the pull P the component of the weight down the slope assist the pull. Resolving down parallel and perpendicular to the plane: P + W sin θ = F = µR N and R N = W cos θ If the pull P is acting up the plane (Fiq. 4.5) and resisting the downward motion of the body, then when the body is just about to move down, Fiq. 4.5 we have, P + F = W sin θ R N = W cos θ and F = µR N
- 6. FRICTION J3010/4/6 When P = 0, the component of the weight down the slope is just able to overcome the friction force F, F = W sin θ and since, R N = W cos θ therefore, F W sin θ = = tan θ R N W cos θ but, F =µ RN thus, tan θ = µ Example 4.2 The pull required to haul a load of 500 kg along a horizontal surface is 1.2 kN. Find the pull parallel to a track of slope 20˚ that is required to haul the load up the inclined plane. Assuming the coefficient of friction to be the same in all cases. Solution 4.2 Fiq. 4.6 Given: M = 500 kg P = 1200 N θ = 20˚ P=F (for horizontal surface) P = 1200 N Resolving parallel to the plane: P = F + W sin θ P = 1200 + 500(9.81) sin 20 0 P = 1200 + 1677.6 P = 2.87 kN
- 7. FRICTION J3010/4/7 Activity 4A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! Fiq. 4.7 4.1 Determine the magnitude of the push P (Fig.4.7) that will have the wooden crate on the point of sliding. Coefficient of friction between the crate and the floor is 0.55. Wooden crate = 20.41 kg. 4.2 A 2 tonne truck is pulled at uniform speed up an inclined of 1 in 10 along the slope by a rope, parallel to the inclined. The rope passes over a frictionless pulley at the top of the inclined and has a 275 kg mass hanging freely at its end. Find the resistance to motion parallel to the inclined. 4.3 A pull of 550 N parallel to an inclined plane is required to just move a mass of 50 kg up the plane. If a pull is applied parallel to the plane to drag the mass down at a steady speed of 4 m/s, what is the power required? The plane is inclined at 30 ° to the horizontal. 4.4 A load of mass 1350 kg lies on a gradient of 60 ° to the horizontal. If static friction µs = 0.5 and kinetic friction, µk = 0.4. Calculate: (i) the pull parallel to the gradient required to prevent the load from sliding down, (ii) the pull required to haul the load up the gradient at uniform speed.
- 8. FRICTION J3010/4/8 Feedback To Activity 4A Have you tried the questions????? If “YES”, check your answers now 4.1 186 N 4.2 736 N 4.3 240 W 4.4 (i) 8.14, (ii) 14.1 kN
- 9. FRICTION J3010/4/9 INPUT 4.3 THE ANGLE OF FRICTION AND TOTAL REACTION Fiq. 4.8 Consider a body that is about to move to the right (Fiq. 4.8). The force R is the resultant of RN and F. Since the latter are at right angles, the angle φ between R and RN is given by, ab tan φ = oa F = RN µR N = RN =µ
- 10. FRICTION J3010/4/10 4.4 APPLICATION OF ANGLE OF FRICTION TO MOTION ON THE INCLINED PLANE. Fiq. 4.9 Fig. 4.9 shows a body being moved up a plane by means of a horizontal pull P. Since motion is up the plane, R must have a component down the plane to provide the resisting friction force and is therefore directed as shown. The pull P, weight W and resultant R form a triangle of forces and since P and W are at right angles, thus P = W tan(θ + φ ) Fiq. 4.10 Figure 4.10 shows the case whose body is just about to move down the plane against a resisting horizontal force P. From the triangle of forces: P = W tan(θ − φ ) In this case the angle of the plane θ is greater than the angle of friction φ .
- 11. FRICTION J3010/4/11 Example 4.3 A casting of mass 2 tonnes is to be pulled up a slope inclined at 30 by a pull at an angle to the slope. If the coefficient of friction is 0.3. Find the least force required and its direction to the horizontal. Solution 4.3 Fiq. 4.10 From the triangle of forces: Hence, the minimum value of P is given by: P = W sin (30 + φ ) where tan φ = 0.3 or φ = 16˚ 42’ W = Mg = 2 x 9.8 = 19.6 kN hence, P = 19.6 sin 46˚ 42’ = 14.25 kN
- 12. FRICTION J3010/4/12 Activity 4B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 4.5 A body of weight of 50 kg is hauled along a rough horizontal plane, by a pull of 18 kg acting at an angle of 14˚ with the horizontal. Find the coefficient of friction. 4.6 Find the horizontal force required to drag a body weighing 10 kg along a horizontal plane, if the plane when gradually raised up to 15˚ the body will begin to slide down. 4.7 A crate with a mass of 50 kg will just slide with uniform speed down a rough ramp at 300 to the horizontal. Find the coefficient of friction. 4.8 A load of mass 1 Mg is to be hauled slowly at constant speed up an inclined of 1 in 2 (sine). If the coefficient of sliding friction is 0.6. What pull parallel to the inclined would required? If the load is placed on a trolley of mass 200 kg what would then be the pull? The coefficient of rolling resistance is 0.12 and friction at the bearing journals may be neglected
- 13. FRICTION J3010/4/13 Feedback to Activity 4B Have you tried the questions????? If “YES”, check your answers now 4.5 0.382 4.6 2.68 kg 4.7 0.576 4.8 10 kN, 7.1 kN.
- 14. FRICTION J3010/4/14 SELF-ASSESSMENT 4 You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment 4 given on the next page. If you face any problems, discuss it with your lecturer. Good luck. Fiq. 4.11 1. The force F (Fig.4.11) just prevents the wooden crate of 100 kg from sliding down the steel chute. Find F if µs for the mating surface = 0.1. 2. A load of 300 kg will just start to slide down a 25o slope. What horizontal force will be required to haul the load up the slope at constant speed? What is the least force required to haul the load up the incline?. State the direction of this least force. 3. A body, resting on a rough horizontal plane, requires a pull of 18 kg inclined at 30 0 to the plane just to remove it. It was found that a push of 22 kg inclined at 30 0 to the plane just removed the body. Determine the weight of the body and the coefficient of friction. 4. Find the force required to move a load of 30 kg up a rough plane, the force being applied parallel to the plane. The inclination of the plane is such that a force of 6 kg, inclined at 30 0, to a similar smooth inclined plane, would keep the same load in equilibrium. The coefficient is 0.3.
- 15. FRICTION J3010/4/15 Feedback To Self-Assessment 4Have you tried the questions????? If “YES”, check your answers now 1. 303 N 2. 3.5 kN, 2.25 kN at 50o to the horizontal 3. 99.4 kg ; 0.1726 4. 14.06 kg. If all your answers are correct, CONGRATULATIONS!!!! ….

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment