PHYSICS 149: Lecture 6

• Chapter 2

– 2.7 Contact Forces: Normal Force and Friction

– 2.8 Tension

Lecture 6 Purdue University, Physics 149 1 ILQ 1

If the distance to the moon were halved, then the force of attraction between the earth and moon would be:

A) quartered (.25 x) B) halved (. 5 x) C) doubled (2 x) D) quadrupled (4 x)

Lecture 6 Purdue University, Physics 149 2 Normal (= Perpendicular) Force • The normal force is a contact force perpendicular to the contact surfaces that prevents two objects from passing through one another. • Normal force is a vector. – Direction: always perpendicular to the “contact surface” (()rather than the horizon) – Magnitude: depends on the weight of the object (see different cases on next pages)

• Type: contact force (not long-range force) • Normal force is usually denoted by N.

Lecture 6 Purdue University, Physics 149 3 Normal Force

• Symbol for FBD: N • Type: Contact force • Direction is normal (perpendicular) to surface • This i s th e norma l componen t f or th e cont act force between two (planar) surfaces

Lecture 6 Purdue University, Physics 149 4 What Causes Normal Force?

• Atoms inside solid objects are inter-connected by molecular bonds which act like springs. • When yypou place an ob ject on to p of a table, the table deforms slightly. This bend is usually not visible to the eye. • The "springs" holding the atoms ithtblin the table compress or s tthtretch exerting a force on the object on the table.

Lecture 6 Purdue University, Physics 149 5 Normal Force: Case 1

• If the table’s surface (contact surface) is horizontal, – Direction of the normal force is perpendicular to the “contact surface.” In this case, vertically upward. – Magnitude of the normal force is the book’s weight, according to Newton’s First Law of Motion.

N = W (= mg) according to

ΣFy = 0 for an object in equilibrium

Lecture 6 Purdue University, Physics 149 6 Normal Force: Case 2

• If the contact surface is horizontal and there is another vertical force acting on the book, – Direction of the normal force is perpendicular to the “contact surface.” In this case, vertically upward. – Magnitude of the normal force is the book’s weight plus the magnitude of the additional force, according to Newton’s First Law of Motion.

N = W (= mg) + F according to

ΣFy = 0 for an object in equilibrium Lecture 6 Purdue University, Physics 149 7 Normal Force: Case 3

• If the contact surface is not horizontal (with an inclination angle θ), – Direction of the normal force is perpendicular to the “contact surface.” In this case, it is not vertical. – Magnitude of the normal force is +y the book’s weight times cosθ, according to Newton’s First Law of Motion. θ N = W cosθ (mg(= mg cosθ) +x Wcosθ according to

ΣFy = 0 for an object in equilibrium Wsinθ θ Lecture 6 Purdue University, Physics 149 8 ILQ 2

A box sitting on a table experiences a normal force N. If you push down on the box, the normal force

A) increases. B) stays the same. C) decreases.

Lecture 6 Purdue University, Physics 149 9 ILQ 3

A box of mass m sits on an inclined plane. What is the relationship between the weight of the box, W, and the magnitude of the normal force exerted on the box, N?

A) W > N B) W = N C) W

Lecture 6 Purdue University, Physics 149 10 ILQ: Net Force

• Compare the net force on the two books.

A) Fphysics > Fbiology B) Fphysics = Fbiology C) Fphysics < Fbiology

Net force is zero on both books! Normal force from table exactly cancels downward forces.

Physics Biology

Lecture 6 Purdue University, Physics 149 11 ILQ: Normal Force Consider a horse pulling a buggy. Is the followinggg statement true? The weight of the horse and the normal force exerted by the ground on the horse constitute an interaction pair that are always equal and opposite according to Newton's third law. A) yes B) no The norma l force is no t an action-reaction force pair of an object ' s weight

Lecture 6 Purdue University, Physics 149 12 G G G G G G G N + W = 0 N + W = 0 N + W + F = 0 In y : N −W = 0 In y : N −W − F = 0 In y: N −W sinθ = 0 N = W N = W + F N = W sinθ

y

Wcos θ θ Wsin θ x

Σ F =0 Equilibrium Σ F=0 x Σ Fy=0 Components of Weight

α+β=90°

=Wx Thus, β=ϕ α+ϕ=90° =Wy

Lecture 6 Purdue University, Physics 149 14 Friction Force • Symbol for FBD: f • Type: Contact force • Direction: Parallel to the surface. This is the parallel component of the contact force between two surfaces •Maggppnitude of frictional force is proportional to the Normal force • Static friction

– fstatic ≤ μs N μs coefficient of Static friction – Direction opposite to sum of other “parallel forces” • Kinetic friction

–frictionfkinetic = μk N μk coefficient of Kinetic friction – Direction opposes motion – Sma ller than s ta tic fr ic tion

• Note: Static friction can be any value up to μsN Lecture 6 Purdue University, Physics 149 15 Friction

• Magnitude of frictional force (llltf)itil(parallel to surfaces) is proportional to the normal force.

– fkinetic = μk N μk coefficient of kinetic friction

–fstatic ≤ μs N μs coefficient of static friction • Be Careful!

– Static friction ≤, can be any value up to μsN – Direction always opposes motion

Lecture 2 Purdue University, Physics 220 16 Friction • Friction is a contact force parallel to the contact surface, and it acts to prevent the objects from slipping on each other. • There are two types of frictions. – Static: the two objects are at rest with respect to each other. – Kinetic: the two objects are slipping one another. • Friction is a vector. – Magnitude is proportional to the magnitude of the normal force. (Note: it does not depend on the contact area. ) – Direction: • Static: the direction that tends to keep the surfaces from beginning to slide. • Kinetic: the direction that would tend to make the sliding stop. • Type: contact force (not long-range force)

• Magnitude of friction is usually denoted by fs (static) or fk (kinetic). Lecture 6 Purdue University, Physics 149 17 Magnitude of Friction

• Static Friction

0 ≤ fs ≤ μsN where μs: coefficient of static friction – Note that μs is a dimensionless constant while the unit for fs and N (normal force) is Newton (N).

– An object starts sliding when an applied force is larger than μsN.

• Kine tic F ri cti on

fk = μkN where μk: coefficient of kinetic friction – Note that μk is a dimensionless constant while the unit for fk and N (normal force) is Newton (N). • For an object on a given surface, μ > μ s k Î fs,max > fk

Lecture 6 Purdue University, Physics 149 18 Friction vs. Applied Force

• At rest, the magnitude of static friction (fs) is equal to the maggpp(nitude of an applied force (Fapp).

• When an applied force becomes larger than fs,max (= μs N), an object starts sliding. • Once the object starts sliding, a smaller force is required in order to keep the object moving at a constant velocity.

– In other words,,q in equilibrium , Fapp has to be fk (= μkN) that is less than fs,max (= μsN) because of μs > μk .

• (dur ing s liding ) If Fapp > fk, the o bject will be acce lera te d.

• (during sliding) If Fapp < fk, the object will stop.

Lecture 6 Purdue University, Physics 149 19 ILQ

To make an object start moving on a surface with friction requires:

A) Less force than to keep it moving on the surface at a constant velocity B) The same force as to keep it moving on the surface at a constant velocity C) A force equal to the weight of the object D) More force than to keep it moving on the surface at a constant velocity

Lecture 6 Purdue University, Physics 149 20 Book Pushed Across Table • ClltfCalculate force o fhdtkf hand to keep th thbklidite book sliding at a const ttant spee difthd, if the

mass of the book is 1 kg, μs = .84 and μk=.75. x-direction: ΣFF0=0 Constant Speed ⇒ΣF=0 Combine: Fhand-Ffriction = 0 Fhand=Ffriction Fhand = μk FNormal Fhand=μk FNormal Fhand=0.75×9.8 N y-direction: ΣF=0 Fhand=7.3 Νewtons F -F = 0 Normal Gravity Normal FNormal = FGravity FNormal =1×9.8=9.8 N friction hand Physics y x Gravity Lecture 2 Purdue University, Physics 220 21 ILQ

A box of weight 50 N is at rest on a floor where

μs = 0. 3. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?

A))toteet to the left B) to the right = 30 N, C) the box does not move right

Lecture 6 Purdue University, Physics 149 22 ILQ Your car won't start, so you are pushing it. You apply a horizontal force of 300 N to the car, but it doesn't budge. Which of these other forces must have a magnitude of 300 N? (a) the frictional force exerted by the road on the car (b) the force exerted by the car on you (c) the frictional force exerted on you by the road

A) (a), (b), and (c) B) (a ) an d (c ) bu t no t (b) C) (a) D) (b) E) (c) Lecture 6 Purdue University, Physics 149 23 What Does Cause Friction?

• Friction is caused by atomic or molecular bonds between the high points on the surfaces of the two object.

• These bonds are formed by microscopic electromagnetic forces that hold the atoms or molecules together. New Bond

Lecture 6 Purdue University, Physics 149 24 Classical Friction

• Conjectures of classical friction: – Friction is proportional to the normal force – Depends on the nature of the surfaces – Does not depend on the area of contact – Friction is independent of velocity

• Friction is much more complicated – Friction may decrease with roughness and in the other extreme very smooth surfaces may have enormous frictional forces (cold welding).

Lecture 6 Purdue University, Physics 149 25 Atomic Friction

• Friction occurs when atoms close to one surface are set in motion byyg the sliding action of atoms in the other surface. • These vibrations (phonons) are sound wav es w hich get conv erted into heat. • The amount of mechanical energy (the energy needed to keeppj the object movin g)pg) converted to phonons de pends on the sliding substance because different solids vibrate at different frequencies. If one solid has many of the same frequencies as the other then friction will be high. • Friction can depend on the contact area of the two objects. Contact area is the area that actually touches the other object. The more contact area the higher the friction.

Lecture 6 Purdue University, Physics 149 26 Example: Inclined Plane A 1.0 kg block sits on an inclined plane which has an angle with the horizontal of 25 degrees. What is the force of st ati c f ri cti on? x-component: At rest (right before sliding) Î W sinθ – fs,max = 0 equilibrium Î net force = 0 y

Thus, fs,max = W sinθ N = mg sinθ f =1kgx98m/s= 1 kg x 9.8 m/s2 xsin25x sin 25° = 4.14 N y-component: θθ N – W cosθ = 0 By definition, f = μ N x s,max s W Thus, μs = fs,max/N = (W sinθ) / (W cosθ) = tanθ = tan 25° = 0.47 Lecture 6 Purdue University, Physics 149 27 Tension • At any point in the rope (or string, cable or chain), tension is the pulling force exerted on the rope on one side of the point by the rope on the other side.

• At its two ends, tension is the pulling force exerted on the object attached to its ends by the ropes at the ends.

• Note that tension can pull but not push.

=T1 “If” the chain’s weight is not negligible, •=T4 =T2 T1 > T2 > T3 > T4 .

=T 3 FlFor example,

T1 = T4 + chain’s weight. Lecture 6 Purdue University, Physics 149 28