TOPIC - FRICTION

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.force

The constant ratio which the limiting friction bears to the normal reaction is called CO-EFFICIENT OF FRICTION. It is usually denoted by the letter μ Thus, F=μR where F= limiting friction R=normal reaction

The limiting friction F and the normal reaction R acting at right angles to each others have a resultant force say S, which is called the RESULTANT REACTION. Thus S =√(R²+F²) = √(R²+μ²R²) = R√(1+μ²)

When a body is in limiting equilibrium on another, the angle which the resultant reaction S at the point of contact makes with the normal reaction is called the ANGLE OF FRICTION and is generally denoted by the greek letterλ. Tanλ=μ Hence the tangent of the angle of friction is equal to the co-efficient of friction

F R S λ Tanλ = μ Where F= limiting friction R= normal reaction

Let F be the limiting friction, R be the normal reaction and S be the resultant reaction. If λ be the angle of friction which S makes with R, then S cosλ = R ……………(1) S sinλ = F ……………(2) Dividing (2) by (1), we have tan λ = F / R = µ Hence the tangent of angle of friction is equal to the co-efficient of friction.

The cone of friction is the cone which has the point of contact as its vertex, the normal as its axis and λ as its semi vertical angle

R Cone of friction Point of contact θ Axis of friction

A heavy body is placed on a rough inclined plane of inclination ‘α’ greater than the angle of friction, being acted upon by a force parallel to the plane and alone a line of greater slope. To find the limits between which the force must lie.

Case-1:Let the body be on the point of moving up the plane Let P1 be the force acting parallel to the plane keeping the body at rest so that the force of friction R acts down the plane Resolving along and perpendicular to the plane P1=μR+Wsinα…………………(1) R=Wcosα…………………………..(2) Eliminating R from (1) and (2)

Body moving upward O μRμR α α

P1=W(sinα+μcosα) =W(sinα +(sinλ/cosλ)cosα) on solving P1=Wsin(α+λ)/cosλ Which gives the amount of force.

CASE 2:Let the body be on the point of moving down the plane Let P2 be the force required to keep the body at rest since body is on the point of moving down the plane the force of friction acts up the plane Resolving along and perpendicular to the plane P2+μR=Wsinα R=Wcosα P2=Wsinα-Wcosα =Wsin(α-λ)/cosλ

Body moving down the plane P2 α α μRμR

To find the limits between which a force must lie in order to keep a body in equ. on rough inclined plane when the force acts horizontally : Let ‘α’be the inclination of the plane to the horizontal ‘W’ be the weight of the body and r the normal reaction. Let’μ’ be the co-efficient of the friction and ‘λ’be the angle of friction’

CASE 1 : LET THE BODY BE ON THE POINT OF MOVING UP THE PLANE Let P1 be the horizontal force required to keep the body at rest so that the force of friction μR acts down the plane

μRμR o B α α α

Resolving the force acting on the body along and perpendicular to the plane P1cosα=μR+W sinα………………………..(1) R=W cos α+P1sinα…………………..(2) Eliminating R from (1) and (2) P1cosα =μ(W cos α+P1sinα)+W sinα P1(cos α-tanλsinα)=W(tanλ cosα+sinα) On solving P1=W tan(α+λ)

CASE 2:LET THE BODY BE ON THE POINT OF MOVING DOWN THE PLANE Let P2 be the force required to keep the body at rest since the body is on the point of moving down the plane the force of friction acts up the plane

μRμR o α α α

Resolve the force along and perpendicular P2 cos α+μR=Wsinα…………………..(3) R=wcosα+P2sinα……………………….(4) Eliminating R from (3) and (4) P2cos α+μ (Wcosα+P2sinα)= Wsinα P2(cosα+tanλ.sinα)=W(sinα-tanλ.cosα) On solving P2=Wtan(α-λ)

EXAMPLE TWO EQUAL WEIGHTS ARE ATTACHED TO THE ENDS OF THE STRING WHICH IS LAID OVER THE TOP OF TWO EQUALY ROUGH PLANES HAVING THE SAME ALTITUDES AND PLACED BACK TO BACK. THE ANGLES OF THE INCLINATION OF THE PLANES TO THE HORIZON BEING 30 ˚AND 60˚ RESP.SHOW THAT THE WEIGHTS WILL BE ON THE POINT OF MOTION IF THE COEFFICIENT OF FRICTION BE 2-√3

SOLUTION Let R and R’ be the planes with the inclinations of 60˚ and 30˚ resp. and T be the tension of string The weights W on the plane of inclination 60˚ is on the point of moving downwards therefore the friction μR on this plane acts up the plane and the friction μR’ on the other plane acts down the plane.

R R’ T T μRμR W W μR’

Resolving along and perpendicular to the plane (i) T+μR=Wsin 60˚ R=Wcos60˚ T=Wsin60˚-μR =Wsin60˚-μWcos60˚ =W(sin60˚-μcos60˚) (ii) T-μR’=Wsin30˚ R’=Wcos30˚ T=W(sin30˚+μcos30˚)

Equating the two values of T we get W(sin60˚-cos60˚)=Wsin30˚+cos30˚ μ=(√3-1)/(√3+1) =2-√3

EXAMPLE THE FORCE ACTING PARALLEL TO A ROUGH INCLINED PLANE OF INCLINATION ‘α’TO THE HORIZON JUST SUFFICIENT TO DRAW A WEIGHT UP THE PLANE IS n TIMES THE FORCE WHICH WILL JUST LET IT BE ON THE POINT OF SLIDING DOWN THE PLANE. PROVE THAT tanα=μ(n+1)/(n-1)

SOLUTION: Body moving upward O μRμR α α

Let P1 be the force acting up the plane parallelto the plane which is just sufficient to draw a weight W up the plane P1=Wsin(α+λ)/cosλ……………………..(1) Let P2 be the force which is just sufficient to support the body P2=Wsin(α-λ)/cosλ…………………(2)

Body moving down the plane P2 α α μRμR

Now according to the given condition P1 =n.P2 From (1) and (2) Wsin(α+λ)/cos λ=nWsin(α-λ)/cosλ Sin(α+λ)/sin(α-)λ=n/1 Applying c-d rule, we have (Sin(α+λ)+sin(α-λ))/(sin(α+λ)-sin(α- λ))=(n+1)/(n-1) On solving Tanα=tanλ.(n+1)/(n-1) =μ(n+1)/(n-1)

A uniform rod of length 2l rests in a vertical plane against a smooth horizontal peg at a height h, the lower end of the rod being on level ground. Show that if the rod be on the point of slipping when its inclination to the horizontal is θ,then the co=efficient of friction between the rod and the ground is L sinθ sin2θ/(2h-lcosθ sin2θ)

Solution Let AB be the rod resting with the end A on the level ground and a point of its lenth resting over the peg P. let the normal reaction at A and P be R and S resp. and R the force of friction. Resolving vertically and horizontally

θ

R+Scosθ=W And S sinθ=μW S=μW/(sinθ+μ cosθ) Taking moments about A W l cos θ=S h cosecθ =(W/(sinθ+μ cosθ))*hcosecθ μ(h-lsinθcos²θ)=lsin²θcosθ hence=lsinθ cos2θ/(2h-lcosθsinθ)

ASSIGNMENT

1. A weight of 60 kg can just rest on a rough inclined plane of inclination 30˚ to the horizon,when the inclination is increased to 60˚ find the least horizontal force which will support it ? 2. Find how high can a particle rest inside a hollow sphere of radius ‘a’ if the co-efficient of friction be 1/√3? 3. A weight can be just supported on a rough inclined plane by a force P acting along the plane or by a force Q acting horizontaly show that weight is PQ/√(Q²sec²ø- P²) where ‘ø ‘ is the angle of friction?

4.A body of weight 80 kg rest on rough horizontal plane while a force of 20 kg is acting on it in a direction making an angle of 60˚ with the horizontal.find the force of friction that is called into play ? 5.A heavy body is placed on a rough inclined plane to find the force just sufficient to move the body up the plane the force acting in a vertical plane through the line of greatest slope to the body. 6.A uniform rod rest with one extremity against a rough vertical wall the other being supported by a string of equal length fastened to point in the wall. Prove that the least angle which the string can make with the wall is tan ¹ ־ 3/μ?

TEST NOTE : do any three.

1.To find the limits between which a force must lie in order to keep a body in equilibrium on a rough inclined plane when the force acts horizontally? 2. Find how high can a particle rest inside a hollow sphere of radius ‘a’ if the co-efficient of friction be 1/√3? 3.A uniform rod rest with one extremity against a rough vertical wall the other being supported by a string of equal length fastened to point in the wall. Prove that the least angle which the string can make with the wall is tan ־ ¹3/μ?

4.A ladder inclined at 60˚ to the horizon rest between a rough floor and a smooth vertical wall. Show that if the ladder begins to slide down when a man has ascended so that his center of gravity is half way up.co-efficient of friction between the foot of ladder and this floor is √3/6? 5. A uniform ladder rests in limiting equilibrium with one end on a rough floor whose co-efficient of friction is and with the other end against a smooth vertical wall. Show that the inclination to the vertical Is tan ־ ¹2μ?

THANKS