Chapter 4. Forces and Newton s Laws of Motion. F=ma; gravity

Chapter 4 Forces and Newton s Laws of Motion F=ma; gravity

0) Background Galileo inertia (horizontal motion) constant acceleration (vertical motion) Descartes & Huygens Conservation of momentum: mass x velocity = constant Kepler & Braha laws of planetary motion (kinematics only) Question of the day: Explain planetary motion

1) Newton s first law: the law of inertia A free object moves with constant velocity No forces at rest or motion in a straight line Natural state is motion with constant velocity - Aristotle: rest is natural state - Galileo: circular motion (orbits) is natural state

Inertial reference frame - A reference frame in which the law of inertia holds - Requires ability to identify a free object: If no force acts on a body, a reference frame in which it has no acceleration is an inertial frame. - does not hold on a carousal, or an accelerating car a

Velocity is relative...have the ship proceed with any speed you like, so long as the motion is uniform and not fluctuating this way and that. You will discover not the least change in all the effects named, nor could you tell from any of them whether the ship was moving or standing still. from Dialogue Concerning the Two Chief World Systems, 1632.

Galileo's Principle of Relativity The mechanical laws of physics are the same for every inertial observer. By observing the outcome of mechanical experiments, one cannot distinguish a state of rest from a state of constant velocity. Absolute rest cannot be defined.

A cup of coffee is sitting on a table in a recreational vehicle (RV), and then it slides toward the rear of the RV. According to Newton s first law, which one or more of the following statements could describe the motion of the RV? 1. The RV is at rest, and the driver suddenly accelerates. 2. The RV is moving forward with constant velocity, and then accelerates. 3. The RV is moving backward with constant velocity, and then the driver hits the brakes (decelerates). A.1 B.1 & 2 C.3 D.none E.1, 2, & 3

A pipe is bent into the shape shown and oriented so that it is sitting horizontally on a table top. You are looking at the pipe from above. The interior of the pipe is smooth. A marble is shot into one end and exits the other end. Which one of the paths shown in the drawing will the marble follow when it leaves the pipe? A.1 B.2 C.3 D.4 E.5

2) Newton s second law: F=ma a) Mass - quantity of matter - quantity that determines weight (gravitational mass) (determined with a balance) - quantity that resists acceleration (inertial mass) (i) Define 1 kg as mass of a standard cylinder (ii) Addition of masses (scalar): m = m 1 + m 2 - in particular two identical masses have twice the mass, to satisfy quantity of matter definition (iii) Observe acceleration vs mass for a given force: mass acceleration 1 kg 1 m/s 2 2 kg 1/2 m/s 2 3 kg 1/3 m/s 2 mass is inversely proportional to acceleration

b) Force - push or pull - disturbs natural state: causes acceleration (i) Define 1 N (newton) as force required to accelerate 1 kg by 1 m/s 2 (ii) Addition of forces (vector):! F net =! F 1 +! F 2 +...! F net =! F Identical forces in opposite direction produce no acceleration Two identical forces at 60º produce the same acceleration as a third identical force at 0º (cos(60º)=1/2) Two identical parallel forces corresponds to twice the force.

(iii) Observe acceleration vs. force for a given object Force Acceleration 1 N 1 m/s 2 2 N 2 m/s 2 3 N 3 m/s 2 Force is proportional to acceleration (iv) Types of force: - gravity - electromagnetic - weak nuclear -strong nuclear electroweak.

c) Second Law F a m 1 a (const mass) (const force)} F ma F = (const)ma Define const = 1! F = m! a (F is net force) For m = 1 kg, and a = 1 m/s 2, F = 1 N by definition, so 1 N = 1 kg m/s 2

F = ma can be used as the defining equation for force and inertial mass, but only because of the physical observation that force is proportional to acceleration (for a given mass), and mass is inversely proportional to acceleration (for a given force). Inertia is the tendency of an object not to accelerate Newton s second law formally refers to the rate of change of momentum: For constant mass,! F = Δ(m! v) Δt! F = m Δ! v Δt = m! a Special case:! F = 0 0 = m! a! a = 0 velocity is constant (1st law)

d) Free-body diagrams and net force Replace object(s) by dot(s). Represent all forces from the dot. Solve F=ma for each object F 2 F N F 2 F 1 m F 1! F = net! F mg Net force is vector sum of all forces

10 N? N m m m If the scale on the left reads 10 N, what does the scale on the right read? A. 0 N B. 10 N C. 20 N

10 N 10 N 10 N scale? N m 10 N 10 N scale m m

Example CJ 4.1 If the force on an airplane is 3.7 x 10 4 N, the mass of the plane is 3.1 x 10 4 kg, and the mass of the pilot is 78 kg, what is the force on the pilot?

Example Two forces FA and FB are applied to an object whose mass is 8.0 kg. The larger force is FA. When both forces point due east, the object s acceleration has a magnitude of 0.50 m/s 2. However, when FA points due east and FB points due west, the acceleration is 0.40 m/s 2, due east. Find (a) the magnitude of FA and (b) the magnitude of FB.

e) Components of force! F = m a! means F x = ma x & F y = ma y sum of all forces

Example: m = 1300 kg m θ = 67º F 1 = 15 N F 2 = 17 N y Find acceleration. F 1 θ F 2 x F x = F 2 + F 1 cosθ = 23 N F y = F 1 sinθ = 14 N a x = a y = F x m F y m = 0.018 m/s2 = 0.011 m/s2

Problem solving with F=ma Choose an inertial reference frame earth is close enough measure v and a relative to this frame Draw free-body diagrams for each object Normal Tension Friction Velocity Weight Indicate direction of velocity and acceleration

Normal F N T Tension Friction f θ Velocity W Weight Resolve forces and add components F y = F N W + T sinθ F x = T cosθ f

Apply 2nd law F x = ma x F y = ma y Use kinematic equations for constant accel. if force is const.

A car of mass m is moving at a speed 3v in the left lane. In the right lane, a truck of mass 3m is moving at a speed v. As the car is passing the truck, the traffic light ahead turns yellow. Both drivers apply the brakes. What is the ratio of the force required to stop the truck to that required to stop the car? Assume each vehicle stops with a constant deceleration and stops in the same distance x. A. 1/9 B. 1/3 C. 1 D. 3 E. 9

3) Newton s third law For every action, there is an equal and opposite reaction F BA Force on B due to A A B F AB F AB = -F BA Conservation of momentum:! F AB = Δ(m! Av A ) Δt! F BA = Δ(m! Bv B ) Δt! F AB + F! BA = Δ(m! Av A ) + Δ(m! Bv B ) Δt Δt 0 = Δ(m!! Av A + m B vb ) Δt Change in momentum of an isolated system is zero

A chain of 3 links, each having a mass of 100 g, is pulled on a surface with an acceleration of 2.5 m/s 2. Find the forces acting between adjacent links.

Two ice skaters, Paul and Tom, are each holding on to opposite ends of the same rope. Each pulls the other towards him. The magnitude of Paul s acceleration is 1.25 times greater than the magnitude of Tom s acceleration. What is the ratio of Paul s mass to Tom s mass? (a) 0.67 (b) 0.80 (c) 0.25 (d) 1.25 (e) 0.50

Homework C&J 4.14 A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the ball is 0.38 kg. The ball approaches the cushion with a velocity of 2.1 m/s and rebounds with a velocity of 2.0 m/s. The ball remains in contact with the cushion for a time of 3.3 ms. What is the average net force (magnitude and direction) exerted on the ball by the cushion?

4) Gravity Newton s law of universal gravitation: G = 6.673 10 11 Nm 2 /kg 2

Spheres look like points from the outside

Gravitation: A very special force. gravitational masses inertial mass F = Gm g M r 2 F = m i a If m i = m g, F a = Fr2 GM a = GM r 2 Acceleration independent of mass = g equal to at least ppt

It is, however, clear that science is fully justified in assigning such a numerical equality only after this numerical equality is reduced to an equality of the real nature of the two concepts. -- Einstein

The principle of equivalence A uniform gravitational field is completely equivalent to a uniformly accelerated reference frame. No local experiment can distinguish them Concept of inertial frame no longer useful Gravity is geometrical

Weight, W force due to earth s (or a planet s) gravity r E = 6.38 10 6 m M e = 5.98 10 24 kg

Near the surface, r = R E W = G M E R E 2 m or, W = mg where g = GM E R E 2 = 9.8m/s 2

Recently, scientists reported the discovery of a new extra-solar planet: Gliese 581g. Gliese 581g is interesting because its surface temperature may allow liquid water and therefore life to exist. Gliese 581g has a mass 3 to 4 times that of earth and the estimated diameter is 1.2 to 1.4 times that of earth. Estimate the acceleration due to gravity on the surface of the new planet. (Use 4 and 1.4) A.9.8 m/s 2 B. 33 m/s 2 C. 20 m/s 2 D.6.4 m/s 2

5) Normal Force Perpendicular force of surface on object constraint F y = 0 F N W = 0 F N = W

F y = 0 F N 11N W = 0 F N = W +11N = 26N Normal force increased

F y = 0 F N +11N W = 0 F N = W 11N = 4N Normal force decreased

Perpendicular to surface (not necessarily up)

Apparent weight Apparent weight (measured by scale) is the normal force At rest or moving with constant velocity F N W F N W = ma = 0 F N = W = 700N

Accelerating up F N W a = F N W m F N W = ma y = ma F N = W + ma = 1000N = F N W W g = F N W 1 g If F N = 2W, a = g For F N = 1000N, and W = 700N, a = 3 7 g

Accelerating down F N W F N W = ma y = ma F N = W ma = 400N a = W F N m = W F N W g = 1 F N W g If F N = 0, a = g For F N = 400N, and W = 700N, a = 3 7 g

Free fall F N = 0 W F N W = ma y = ma = mg F N = W mg = 0 weightlessness

6) Friction Parallel force of surface on object (between surfaces) Resists motion opposite to direction of motion or applied force (<= applied force; does not cause motion) Material dependent

Kinetic friction opposes sliding motion f k = µ k F N coefficient of kinetic friction

Static friction constraint f s µ s F N coefficient of static friction f s MAX = µ s F N f s = F if F f s MAX

Three identical blocks are being pulled or pushed across a horizontal surface by a force, as shown in the drawings. The force in each case has the same magnitude. Rank the kinetic frictional forces that act on the blocks from highest to lowest. a) A > C > B b) B > A > C c) C > A > B d) A > B > C e) A = B = C

A force F is applied to three identical blocks. The blocks remain at rest. The force in each case has the same magnitude. Rank the static frictional forces that act on the blocks from highest to lowest. a) A = B = C b) A > B = C c) C > A > B d) C > B > A e) B > A = C

Example C&J 4.99 A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs 31 N. The coefficient of static friction between his hands and the book is 0.40. To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

Forces on an inclined plane F N y f k motion m θ x θ mg F = F mgcosθ = 0 y N F x = mgsinθ f k = ma x 51

Oct. 2005 Midterm test Q7 A 1.0 kg mass is released at the top of an inclined plane that is 5.0 m long and makes an angle of 32 above the horizontal. If the coefficient of friction is 0.20, what is the speed of the object when it reaches the bottom?

7) Tension and pulleys Tension: force exerted by rope or cable For an ideal line, the same force is exerted at both ends objects connected by a taut line have the same acceleration Pulley: changes direction of force For an ideal pulley, the magnitude of the tension is the same on both sides magnitude of acceleration of connected objects is the same

T 1 m 1 m 2 T 2 +a T 1 = T 2 = T a 1 = a 2 = a For the example, a 1y = -a 2y Simplify problem, by choosing sign for a sense of the motion

T T T 1 +a m 1 g m 2 g m 1 m 2 T 2 Equations of motion: T m 1 g = m 1 a m 2 g T = m 2 a Solve for a (eliminate T ): a = m 2 m 1 m 1 + m 2 g Solve for T (eliminate a): 2m T = 1 m 2 m 1 + m 2 g

+a Acceleration can be determined by considering external forces (tension is an internal force holding objects together) ( ) F ext (+) = m a + m 1 m 2 g m 1 g = (m 1 + m 2 )a m 1 g m 2 m 2 g a = m 2 m 1 m 1 + m 2 g

If m 1 = m 2, and rope and pulley are ideal, what happens when the monkey climbs the rope? m 1 A) bananas move up, monkey stays B) monkey moves up, bananas stay C) monkey moves up slower than the bananas D) monkey moves up faster than the bananas E) both move up at the same speed m 2

If m 1 = m 2, and rope and pulley are ideal, what happens when the monkey climbs the rope? T 1 T 2 m 1 T 1 T 2 m 2 m 1 g m 2 g Since T 1 = T 2, any change in T 2 to cause the monkey to ascend, results in a change in T 1, causing the bananas to ascend at the same rate.

If m 1 = m 2, and rope and pulley are ideal, what happens when the monkey climbs the rope? T 1 T 2 m 1 g m 2 g Since T 1 = T 2, any change in T 2 to cause the monkey to ascend, results in a change in T 1, causing the bananas to ascend at the same rate.

8) Equilibrium applications Equilibrium means zero acceleration Balance forces in x and y directions F x = 0 F y = 0

Example: Find tension on leg (F) Free body diagram for pulley: T Free body diagram for weight: T mg T=mg Since T = T 1 = T 2, F = T 1 cos 35º +T 2 cos 35º F = 2T cos 35º = 36 N

A block of mass M is hung by ropes as shown. The system is in equilibrium. The point O represents the knot, the junction of the three ropes. Which of the following statements is true concerning the magnitudes of the three forces in equilibrium? A. F 1 + F 2 = F 3 θ = 30 B. F 1 = F 2 = 0.5 F 3 C. F 1 = F 2 = F 3 D. F 1 > F 3 E. F 2 < F 3

9) Non-equilibrium applications Non-equilibrium means non-zero acceleration Determine acceleration from 2nd law: F x = ma x F y = ma y Solve kinematic equations

A sphere of mass 3.0x10-4 kg is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes a constant angle of 37 with the vertical. Find (a) the push magnitude (b) the tension in the cord.

Final Exam Dec. 2005 Q24 A block is at rest on a rough inclined plane and is connected to an object with the same mass as shown. The rope is massless and the pulley frictionless. The coefficient of static friction between the block and the plane is μs. What is the magnitude of the static frictional force acting on the block?

C&J 4-98 Two forces (in addition to gravity and the normal force) act on the 7.00-kg block shown in the drawing. The magnitudes of the forces are F 1 = 59.0 N and F 2 = 33.0 N. What is the horizontal acceleration (magnitude and direction) of the block?

The drawing shows a large cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (mass = 4.0 kg) is in contact with the front surface of the large cube and will slide downward unless P is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that P can have in order to keep the small cube from sliding downward?

What is the direction of the frictional force on the smaller block? A. up B. down C. to the right D. to the left E. There is no frictional force

What force causes the small block to accelerate to the right? A. Gravity B. Friction C. The normal force D. It doesn t accelerate

Homework C&J 4.87 The alarm at a fire station rings and an 86-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.0 m). Just before landing, his speed is 1.4 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?