PHYSICS 149: Lecture 5

• Chapter 2

– 2.5 Newton’s Third Law – 2.6 Gravitational Forces – 2.7 Contact Forces: Normal Force and Friction

Lecture 5 Purdue University, Physics 149 1 Newton’s Third Law • All forces come in pairs • Third law forces involve TWO OBJECTS . • The two forces are: – thfthe force obj bjtect one exer ts on obj bjttect two – the force object two exerts on object one • Three ways to st at e th e 3 rd l aw: – Forces on each other are equal and opposite – For every action there is an equal and opposite reaction – You can’t push on something without it pushing back on you

Lecture 5 Purdue University, Physics 149 2 Newton’s Third Law of Motion

• In an interaction between two objects, each object exerts a force on the other. These two forces are equal in magnitude and opposite in direction. – To every action, there is always opposed an equal reaction. – Forces always come in equal but opposite action- reaction pair. • Note that these two forces act on different objects; they do not cancel in any way. • Don’ t forget that forces always exist in pairs.

Lecture 5 Purdue University, Physics 149 3 Free Body Diagram (FBD)

• A simplified sketch of a single object with force vectors drawn to represent every force acting “on” that object. (It must not include any forces that act on other objj)ects.) • FBD is useful to find the net force acting on an object.

Lecture 5 Purdue University, Physics 149 4 Internal and External Forces

• Internal Forces: Forces which act on one part of an object by another part of the same object • External Forces: Forces which act on an object by some other object.

• Net force on a system = vector sum of internal forces + vector sum of external forces • BtBut, vect or sum of fit internal lf forces i s zero b ecause, f rom Newton’s third law, internal forces will occur in equal and opposite pairs and so they contribute nothing to the sum. They never i n fluence t he system ’s mot ion. • Eventually, net force on a system = vector sum of external forces onlyyy. We need to consider external forces only in order to describe the motion of the system.

Lecture 5 Purdue University, Physics 149 5 Examples

• Net force on a baseball = External forces: interaction with the Earth (gravity) what we + interaction with a bat need to + i n teracti on w ith the a iir consider + interactions among protons, Internal Forces: their vector sum neutrons, andld electrons iiin it is zero

• I am hit by myself (internal forces) and other person (external forces). I am pushed due to external forces only. Internal forces do not make any contribution. Lecture 5 Purdue University, Physics 149 6 Pushing a Stalled Car Two people are pushing a stalled car. The mass of the car is 2000kg. One person applied a force of 300 N, the other 400N. Friction opposes this motion with a force of 600N. What is the acceleration of the car:

y

Fc,man1 Fcg=f x Fc,man2

ΣF = +600N − 300N − 400N ΣF 100N m a = = − = −0.05 2 ΣF = −100N m 2000kg s Lecture 5 Purdue University, Physics 149 7 A FBD for Every Situation A force is applied to the right A skydiver is to drag a sled across ddidescending llloosely-packdked snow w ith a with a constant rightward acceleration. veeoctylocity. A football is moving upwards towards its peak after A car is coasting to the having been right and slowing down booted by the punter.

Lecture 5 Purdue University, Physics 149 8 ILQ 1

When a car accelerates from rest, what force causes the acceleration of the car?

A) The rotating engine on the drive shaft B) Th e f orce of th e axel on the tires C) The friction force of the road on the tires

Lecture 5 Purdue University, Physics 149 9 ILQ 2

A person is standing on a bathroom scale. Which of these is not a force exerted on the scale?

A) a contact force due to the feet of the person B) the weight of the person C) a contact force due to the floor D) the weight of the scale

Lecture 5 Purdue University, Physics 149 10 Newton’s Law of Universal Gravitation

F12 F21 m m 1 r 2 • The magnitude of gravitational force is:

(= F12 = F21)

where G = 6.674 × 10-11 Nm2/kg2 (universal gravitational constant)

Note: m1 and m2 need to be in kg, and r needs to be in m. • The directi on ofittilfif gravitational force is: – each object is pulled toward the other’s center (attractive force) – on line connectingg;y the masses; always attractive • very weak, but this holds the universe together! Lecture 5 Purdue University, Physics 149 11 Comparison with EM Force F F2,1 1,2 q2 q1 r12

F1,2 = force on q1 due to q2 = kq 1q 2 2 = F2,1 = force on q2 due to q1 r12 Direction: on line connecting the masses; can be attractive or repulsive k = universal constant = 8.99 x 109 N-m2/c2 F(2 electrons ) 2 gravity Gme −43 ==×2 2.4 10 Eelectric(2 electrons ) kq e −−931 qCmKgee=×1.6 10 = 9.11 × 10 Lecture 5 Purdue University, Physics 149 12 Weight • Weight is the force of gravity on an object with mass • Units of weight are Newtons or Pounds SbtDifftiht!Same mass but Different weight! mM F = G planet r2

r

On earth W=mg where g=9.8 m/s2 Lecture 5 Purdue University, Physics 149 13 Weight on Earth

• Your weight on Earth is the magnitude of Earth’s gravitational force exerted on you (m).

GM m ⎛ GM ⎞ where R is the distance W = E = m⎜ E ⎟ R2 ⎝ R2 ⎠ between you and Earth’ s center • The weight of an object of mass m “near” Earth’s surface is:

where (g is called the gravitational field strength)

Lecture 5 Purdue University, Physics 149 14 Weight on Other Planets

• The weight of an object of mass m “near” a planet’ sssurface surface is:

GM m ⎛ GM ⎞ W = Planet = m⎜ Planet ⎟ = mg at Planet 2 ⎜ 2 ⎟ Planet RPlanet ⎝ RPlanet ⎠

GM Planet where gPlanet = 2 RPlanet

• For example, gMoon = 1.62 N/kg ≈ 1/6 gEarth. – Let’ s say there is a man whose mass is 100 kg. • At the surface of Earth, his mass and weight are 100 kg and

980 N (=m·gEarth), respectively. • At th e sur face o f Moon, his mass an d we ig ht are 100 kg an d

162 N (=m·gMoon), respectively. Lecture 5 Purdue University, Physics 149 15 Weight on Earth and on Moon • How far above the surface of the Earth does an object have to be to have the same weight as it would have on the sur face of th e moon? N eg lec t e ffect s f rom th e Earth’s gravity on the Moon’s surface and vice versa

M E M M FE = Gm 2 = Gm 2 = FM r rM

22M E rr= M M M

24 5.97× 10 Kg 2 r =××=×(1.7410 4 1034kkkm) 1.57 10 km 7.35× 1022 Kg 433 Heightoversurface r-rE =× 1.57 10km − 6.371 × 10 km =× 9.3 10 km

Lecture 5 Purdue University, Physics 149 16 Things are different on the Moon Earth Moon Surface ggyravity 10.17 compared to Earth Your mass 40 Kg 40 Kg Energy to stop a 1 Kg 625 Joules 625 Joules ball moving at 90 km/hour How much can you lift 10 kg 60 kg How higgyjph can you jump 20 cm 120 cm How far can you kick a 20 m 120 m ball

Lecture 5 Purdue University, Physics 149 17 Force of Gravity For objects on the surface of the earth: • F = GMm/R2 = m(GM/R2))g = mg •g = GM/R2 = 9.8 N/kg = 9.8 m/s2 What about at the top of Mount Everest? (h=8850m or GM m 29,035 feet FW== E r2 −2 Wh1 2 ⎛⎞ Everest==+=()r 1 Everest Wr2 ⎜⎟ surface ()rh+ Everest ⎝⎠

⎛⎞hEverest −3 ⎜⎟1− 2=− 1 2.76 × 10 =− 1 0.00278 = 0.99722 ⎝⎠r The approximation h 8.850× 103 Everest ==×1. 389 10−3 works well since: r 6.371× 106

Lecture 5 Purdue University, Physics 149 18 r r Your weight 2 1 decreases as your altitude ggpoes up.

Lecture 5 Purdue University, Physics 149 19 Weight The weight (W) of an object is eqqgual to the magnitude of the gravitational force acting on a body of mass m W = mg DiDropping an o bjtbject causes it to accelerate at free-fall acceleration g

Fg = mg

W = Fg Lecture 5 Purdue University, Physics 149 20 ILQ: Gravitation

Does a man weigh more

A)) at the top of Mt. Everest or

B) at the base of the mountain?

Lecture 5 Purdue University, Physics 149 21 Which Forces Enter in a FBD?

Several force must be taken into account: • Gravity • Normal Force • Friction • Push or Pull • Tension

W Gravity: if the sled has a mass m the force diiWdue to gravity is W = mg

Lecture 5 Purdue University, Physics 149 22 Which Forces Enter in a FBD? N • Normal force: always perpendicular the sur face with whi c h a bo dy is in contact. W • Friction: the frictional force is parallel to the surface and it always opposes the direction of motion. • PhPush or pu ll N fP

W

Lecture 5 Purdue University, Physics 149 23 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 5 Purdue University, Physics 149 24 What Causes Normal Force?

• Atoms inside solid objects are inter-connected by molecular bonds which act like springs.

• When you place an object on top oftblthtbldff a table, the table deforms slightly. This bend is usually not visible to the eye.

Lecture 5 Purdue University, Physics 149 25 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 5 Purdue University, Physics 149 26 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 5 Purdue University, Physics 149 27 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 5 Purdue University, Physics 149 28