Mechanics 1 Friction

Friction Friction is a force that opposes motion and occurs when two surfaces are in contact. Friction Friction It will therefore act in the opposite direction to motion.

Limiting Friction F = µR Laws of Friction When two surfaces are in contact, the force of friction opposes the motion. If two surfaces are in equilibrium, the force of friction is just sufficient to prevent motion and can be found by resolving the forces parallel to the surface. Friction force will not be larger than necessary to prevent motion. The size of friction force is limited. Once the limited friction is reached and the acting force is great enough to overcome friction then motion will occur. At such point equilibrium is broken. That is when: Limiting Friction F = µR

The maximum value of friction is calculated by using the formula: Fmax = µ R Reaction Coefficient of friction dependant upon the material of the surface and object. The friction acting on the particle is Fmax when : 1. The particle is in ‘limiting equilibrium’ (on the point of slipping). 2. The particle is moving.

The following equation is used for friction when an object is at rest and is not moving. NOTE : Friction can be less than or equal to it’s maximum value.

The magnitude of the frictional force can change to ensure equilibrium. Fmax As the force pulling the sleigh increases, so will the friction to ensure equilibrium. If the force increases to a greater magnitude than the maximum value of friction then the sleigh will move. (limiting friction)

Friction On A Surface Reaction Force Tension Weight Resolving Vertically, R = mg. Where g = 9.8 m/s2 Resolving Horizontally, F = T for equilibrium at rest. If T>F there will be a resultant force, which will produce an acceleration in the positive direction. Force =ma = T – F according to Newton’s Second Law of Motion. Where F= µR.

Example 1: A horizontal rope is attached to a crate of mass 70 kg at rest on a flat surface. The coefficient of friction between the floor and the crate is 0.6. Find the maximum force that the rope can exert on the crate without moving it. Rope Weight

R = mg R = 70 x 9.8 = 686 N F < µR F < 0.6 x 686 F < 411.6 N Solution: Reaction Given: m = 70 kg and µ = 0.6 F < µR and R = mg Find: F = ? R = ? Friction Rope R = mg R = 70 x 9.8 = 686 N F < µR F < 0.6 x 686 F < 411.6 N Weight

Example 2: A block of mass 15 kg rests on a rough horizontal surface. Coefficient of friction between block and the plane is 0.25. Calculate the friction force acting on the block when a horizontal force of 50 N acts on the block. State whether the block will move. If it does find its acceleration. W 50 N R F

Solution: Given: m = 15 kg and µ = 0.25 with Horizontal Force = 50 N Using: R = mg R = 15 x 9.8 = 147 N Using: F = µR = 0.25 x 147 = 36.75 N Since F < 50N So the block moves. W 50 N R F Using Newton’s Second Law of Motion: ma = 50 – 36.75 20 x a = 13.25 N Hence, a = 0.66 m/s2

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Friction On A Slope Reaction F mgsinθ mgcosθ Weight Note : On a slope the weight can be split into 2 components perpendicular and parallel to the slope. The perpendicular component of the weight pushes against the surface so it is the Reaction that indicates the push against the surface.

Example 3: 15o The diagram shows an object, of mass 8 kg, on a rough plane inclined at an angle of 15° to the horizontal. (a) Given that the object is at rest, calculate the least possible value of the coefficient of friction. Give your answer correct to two decimal places. (b) Given that the coefficient of friction is 0·1, find the acceleration of the object.

Solution: Given: m = 8 kg and Angle θ = 15o a) Resolving Perpendicular forces: R = mg cos θ = 8 x 9.8 x cos 15o = 75.73N Resolving Parallel forces: F = mg sin θ = 8 x 9.8 x sin 15o = 20.29N Using, F = µR 20.29 = µ 75.73 Hence, µ = 0.27 R mgsinθ F W mgcos θ 15o b) Given: µ = 0.1 F = µR = 0.1 x 75.73 = 7.573N Using F = ma = 20.29 – 7.573 8 x a = 12.717 a = 1.59 m/s2

Example 4: A rough plane is inclined at an angle α to the horizontal where sin α = 3/5. A body of mass 8 kg lies on the plane. The coefficient of friction between the body and the plane is μ. a) Find the normal reaction of the plane on the body. b) The body is on the point of slipping down the plane. Find the value of μ. c) Calculate the magnitude of the force acting along a line of the greatest slope that will move the body up the plane with an acceleration of 0.7 m/s2.

Solution: Given: sin α = 3/5 and m = 8kg α = sin-1 (3/5) = 36.9o a) Resolving Perpendicular forces: R = mg cos θ = 8 x 9.8 x cos 36.9o = 62.72N b) Resolving Parallel forces: F = mg sin θ = 8 x 9.8 x sin 36.9o = 47.04N Using, F = µR 47.04 = µ 62.72 Hence, µ = 0.75 c) Given: a= 0.7 m/s2 F = 8 x 0.7 = T – (47.04 + 47.04) T = 99.68N

Example 5: A particle is held at rest on a rough plane which is inclined to the horizontal at an angle α where tan α = 0.75. The coefficient of friction between the particle and the plane is 0.5. The particle is released and slide down the plane. Find, a) The acceleration of the particle. b) The distance it slides in the first 2 seconds. α

Solution: Given: tan α = 0.75 and µ = 0.5 b) => α = tan-1 (0.75) = 36.87o a) Resolving Perpendicular forces: R = mg cos θ = mgcos 36.87o Resolving Parallel forces: F = mg sin θ = mg sin 36.87o Using, F = µR = 0.5 x mg cos36.87o Using, Newton’s Second Law of Motion: ma = mg sin 36.87o - 0.5 x mg cos36.87o Diving by m on both sides: a = 9.8 x sin 36.87o - 0.5 x 9.8 x cos36.87o a = 1.96 m/s2 b) Given t = 2 seconds Using, S = ut + ½ at2 Since the particle slides from rest. Hence, U = 0 m/s2 S = 0 x t + ½ x 22 x 1.96 = 3.92 m