Forces & Newton's Laws

PHYSICS 1 Forces & Newton’s Laws Advanced Placement

Presenter 2014-2015 Forces & Newton’s Laws

What I Absolutely Have to Know to Survive the AP* Exam

Force is any push or pull. It is a vector. Newton’s Second Law is the workhorse of the AP Physics 1 exam. It allows you to write down mathematical relationships that are true. Thus, for a single body, if you pick any direction and sum up all the positive and negative forces that act on the body along that line, the sum will equal the product of the body’s mass and its acceleration along that line. A Free Body Diagram allows you to identify all of the forces acting on a single body. Neglect one force or add a fictitious force on your FBD and you are in trouble. • Newton’s 1st Law: in an inertial frame of reference, an object in a state of constant velocity (including zero velocity) will continue in that state unless impinged upon by a net external force. If ΣF=0, then a=0 and the object is at rest or moving at a constant velocity in a straight line. The converse is true also, if an object is in a state of constant velocity (including zero velocity) then a=0 and ΣF=0. • Newton’s 2nd Law: A net force acting on a mass causes that mass to accelerate in the direction of the net force. The acceleration (vector) is directly proportional to the net force (vector) acting on the mass and ΣF inversely proportional to the mass of the object being accelerated. a= or ΣFma= m • Newton’s 3rd Law: For every action force, there exists an equal and opposite reaction force. When one object exerts a force on a second object, the second object always responds with a duplicate force in the opposite direction. If you hit a table with your fist, then the size and direction of the force you apply must be equal and opposite the force the table applies to you. Forces are generated in action/reaction pairs that occur on different objects. If you try to apply 800 Newtons of force to a table that can only provide 600 Newtons of reaction force back on you, you will never succeed. The table will break as soon as you exceed 600 Newtons, which is the maximum force it can apply to you.

Key Formulas and Relationships

ΣFF=net =m a

FmgW =

FFstaticfsmax ≤ µ s N ()

FFkineticfk= µ k N () Gm m F = 12 G r2 kgm ΣF = Sum of the forces is the Net Force Newtons (N) = s2 a = acceleration m = mass FW = weight g = acceleration due to gravity

Ffsmax = maximum static frictional force

Ffk = kinetic frictional force

FN = normal force FG = gravitational force r = distance between the centers of two masses

AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not involved in the production of this material. Copyright © 2013 National Math + Science Initiative®, Inc., Dallas, TX. All rights reserved. Forces & Newton’s Laws

Basic Kinds of Forces

Fg = mg

Weight Fg , W Always directed toward the Earth’s center. Force on a free falling body, if we neglect air friction. mm FG= 12 G r 2 Gravitational F A force of attraction between any two massive objects. G When the Earth is one of the two bodies involved, then the force felt by the second body while positioned on the Earth’s surface will always be directed toward the Earth’s center. A force of support, provided to an object by a surface in which the object is in contact. Always directed perpendicular to and away from the surface providing the support.

W In the figure above, a box isN supported by a table. The figure shows all the forces acting on the box and is called a Free Body Diagram (FBD). If

a box, rests on a level table, then the FWmgN == . Notice that the normal force sometimes equals the weight but not always. Normal FN , N

FN y

mgsinθ x mgcosθ θ

If the box is placed on an inclined plane, then the FmgN = cosθ , the component of the weight that is equal and opposite the normal force. For the inclined plane above, the normal force and the weight are not equal and not even in the same direction.

Friction is produced by the atomic interaction between two bodies as they either slide over one another (kinetic friction) or sit motionless in contact with one another (static friction).

FFfsmax ≤ µ s N (static) Static Friction opposes the intended direction of relative sliding. The static frictional force will only be as high as it needs to be to keep the system in equilibrium. If successively greater and greater forces are applied, the static frictional force will counter each push with a force of equal and opposite magnitude until the applied force is great enough to

shear the bonding between the two surfaces. When you calculate fs or

Ffsusing the equation above, you are finding the maximum static frictional force, one of an infinite number of possible frictional forces

that could be exerted between the two bodies. µs is a proportionality constant called the coefficient of static friction. It is the ratio of the static frictional force between the surfaces divided by the normal force acting on the surface.

FFfk= µ k N (kinetic) Kinetic Friction or Dynamic Friction or Sliding Friction is always

Friction Ff , f opposite the direction of motion. The statement that kinetic friction is a function of the normal force only (surface area is independent) is true only when dealing with rigid bodies that are sliding relative to each

other. When you calculate fk or Ff using the equation above, you are finding the single, constant kinetic frictional force that exists between the two bodies sliding relative to one another. No matter their velocity (assuming heating does not alter the coefficient of kinetic friction) the

kinetic frictional force will always be the same. µk is a proportionality constant called the coefficient of kinetic friction. It is the ratio of the kinetic frictional force between the surfaces divided by the normal force acting on the surface

WARNING! The two quantities fs and fk may look the same, but they tell us different things. Kinetic friction is typically less than static friction for the same two surfaces in contact. Note that the normal force sometimes equals the weight but not always. When you draw a free body diagram of forces acting on an object or system of objects, be sure to include the frictional force as opposing the relative motion (or potential for relative motion) of the two surfaces in contact.

FT is a force that is applied to a body by a rope, string, or cable.

FT is applied along the line of the string and away from the body in Tension FT , T question.

Push me, pull you force that does not fall into one of the above categories, for example, a friend shoves you. The magnitude of the applied force is characterized by an F, with any subscript that makes Applied F Subscript sense to solve the problem. Later in the year, you will encounter additional forces, like the electric force, FE, and the magnetic force, FB.

Strategy on Force Problems 1. Take one body in the system and draw a Free Body Diagram (FBD) for it. 2. Choose x and y axes and place them beside your FBD. One axis must be in the direction of the acceleration you are trying to find. If there is no acceleration, then ΣF =0. 3. If there are forces on the FBD that are not along the x and y directions, find their respective x and y components. nd 4. Using Newton’s 2 Law, sum the forces in the x direction and set them equal to max . If a

second equation is needed, sum the forces in the y direction and set them equal to may . 5. Repeat the above process for all the bodies in the system or until you have the same number of equations as unknowns and solve the problem.

Effective Problem Solving Strategies A Free Body Diagram is normally depicted as a box showing all the forces acting on the body. These forces are depicted as arrows. They don’t have to be drawn to scale, but they should have a length that is appropriate for their magnitude. Also, the force vectors do need to be directionally accurate and Free Body Diagram labeled. Do not include components of the force vectors on your FBD. (FBD) When drawing Free Body Diagrams show only the force(s) that act on the body in question and do not show forces that the body applies to other bodies. Also, do not include velocity or acceleration vectors on your free body force diagrams, since you will lose points for extraneous vectors on your FBD.

Example 1 A rope supports an empty bucket of mass 3.0 kg. Determine the tension in the rope when the bucket is (a) at rest and (b) the bucket is accelerated upward at 2.0 m/s2.

Solution Step 1 Draw a Free Body Diagram for each body in the system.

Step 2 Choose x and y axes. In this case we could choose the y axis to be in the same dimension as the tension and the weight.

Step 3 If there are off axes forces, then find the x and y components.

nd Step 4 Using Newton’s 2 law, sum the forces in the y dimension and set equal to may.

(a) When the bucket is at rest, the net force is zero, so that the tension in the rope equals the weight of the bucket.

ΣFyy= ma

FT − mg =0 m F mg3.0 kg ⎛⎞ 10 30N T ==( )⎜⎟2 = ⎝⎠s

(b) When the bucket accelerates upward, the net force is may.

ΣFyy= ma

FT − mg= ma

FT =+= mg ma m( g + a) ⎛⎞mm FkgT =+=3.0⎜⎟ 1022 2.0 36N upward ⎝⎠ss

Example 2

In the diagram below, two bodies of different masses (M1 and M2) are connected by a string which passes over a pulley of negligible mass and friction. What is the magnitude of the acceleration of the system in terms of the given quantities and fundamental constants?

M1 M2

Solution Step 1: Draw a FBD for each body in the system.

There are two forces acting on each of the bodies: weight downward and the tension in the string upward. The tension is distributed throughout the string. The pulley (negligible mass and friction) changes only the direction of motion, not the tension, so the tension is the same on each side of the pulley. Our FBDs should look like this:

T T

Step 2 Choose x and y axes. In this case we could choose the y axis to be in the same dimension as the tension and the weight for both masses.

Step 3 If there are off axes forces, then find the x and y components.

nd Step 4 Using Newton’s 2 law, for each body sum the forces in the y dimension and set equal to may and/or the x dimension. For m1 the tension is positive and the weight is negative since the acceleration is upward.

ΣFmayy=1

FWmaT −11=

For m2 the tension is negative and the weight is positive since the acceleration is downward.

ΣFmayy=2

−FWT +22 = ma

FWmaT −11=

−FWT +22 = ma

WW21−=( mma 12 +)

WW−− mgmg(mmg21− ) a ==21 2 1 = (mm12++) ( mm 12) ( mm 12 +)

Example 3

A block of mass m rests on a horizontal table. A string is tied to the block, passed over a pulley, and another block of mass M is hung on the other end of the string, as shown in the figure below. The coefficient of kinetic friction between block m and the table is μk. Find the magnitude of the acceleration of the system in terms of the given quantities and fundamental constants.

µk

Solution Step 1: Draw a FBD for each body. Note that according to the FBD, the vertical acceleration of the block on the table is zero, since the normal force is directed upward and the weight force is directed downward. There is a horizontal acceleration for the block since the tension is greater than the frictional force.

FN T

W1 W2

Step 2 Choose x and y axes. In this case we could choose the y axis to be in the same dimension as the tension and the weight for both masses. We could choose the x axis to be in the same dimension as the frictional force and the tension pulling the mass to the right.

Step 3 If there are off axes forces, then find the x and y components.

nd Step 4 Using Newton’s 2 law, for each body sum the forces in the y dimension and set equal to may and/or the x dimension.

For block m the net force vertically is 0 since the block is accelerating to the right and not upward nor downward. Summing the forces in the y dimension we find that the normal force equals the weight of the block.

ΣFmayy=

FWN −1 =0

FWmgN ==1

Summing the forces in the horizontal dimension, the direction of the acceleration for block m, we find that

ΣFxx=ma

Tf−k = ma

For mass two, summing the forces in the vertical dimension, the vertical acceleration for block M, we find that

ΣFMyy= a

−TW+2 = Ma

Adding the two equations together we determine the acceleration of the system

Tf−k = ma

−TW+2 = Ma

Wf2 −k =( mMa +) Wf−− Mgmgµ a ==2 kk (mM++) ( mM)

(Mmg− µk ) a = (mM+ )

Since fFmgkkNk==µµ

SECTION A – Linear Dynamics

1. A ball of mass m is suspended from two strings of unequal length as shown above. The magnitudes of the tensions T1 and T2 in the strings must satisfy which of the following relations? (A) Tl = T2 (B) T1 > T2 (C) T1 < T2 (D) Tl + T2 = mg

Questions 2 – 3

A 2-kg block slides down a 30° incline as shown above with an acceleration of 2 m/s2.

2. Which of the following diagrams best represents the gravitational force W. the frictional force f, and the normal force N that act on the block?

(D)

3. Which of the following correctly indicates the magnitudes of the forces acting up and down the incline? (A) 20 N down the plane, 16 N up the plane (B) 4 N down the plane, 4 N up the plane (C) 0 N down the plane, 4 N up the plane (D) 10 N down the plane, 6 N up the plane

4. When the frictionless system shown above is accelerated by an applied force of magnitude the tension in the string between the blocks is (A) F (B) 2/3 F (C) ½ F (D) 1/3 F

5. A ball falls straight down through the air under the influence of gravity. There is a retarding force F on the ball with magnitude given by F = bv, where v is the speed of the ball and b is a positive constant. The ball reaches a terminal velocity after a time t. The magnitude of the acceleration at time t/2 is (A) Increasing (B) Decreasing (C) 10 m/s/s (D) Zero

39 6. A block of weight W is pulled along a horizontal surface at constant speed v by a force F, which acts at an angle of  with the horizontal, as shown above. The normal force exerted on the block by the surface has magnitude (A) greater than W (B) greater than zero but less than W (C) equal to W (D) zero

7. A uniform rope of weight 50 N hangs from a hook as shown above. A box of weight 100 N hangs from the rope. What is the tension in the rope? (A) 75 N throughout the rope (B) 100 N throughout the rope (C) 150 N throughout the rope (D) It varies from 100 N at the bottom of the rope to 150 N at the top.

8. When an object of weight W is suspended from the center of a massless string as shown above, the tension at any point in the string is (A) 2Wcos (B) ½Wcos (D) W/(2cos) (E) W/(cos)

9. A block of mass 3m can move without friction on a horizontal table. This block is attached to another block of mass m by a cord that passes over a frictionless pulley, as shown above. If the masses of the cord and the pulley are negligible, what is the magnitude of the acceleration of the descending block? (A) g/4 (B) g/3 (C) 2g/3 (D) g

40 A plane 5 meters in length is inclined at an angle of 37°, as shown above. A block of weight 20 N is placed at the top of the plane and allowed to slide down.

10. The magnitude of the normal force exerted on the block by the plane is (A) greater than 20 N (B) greater than zero but less than 20 N (C) equal to 20 N (D) zero

11. Multiple correct: Three forces act on an object. If the object is moving to the right in translational equilibrium, which of the following must be true? Select two answers. (A) The vector sum of the three forces must equal zero. (B) All three forces must be parallel. (C) The magnitudes of the three forces must be equal. (D) The object must be moving at a constant speed.

12. For which of the following motions of an object must the acceleration always be zero? (A) Any motion in a straight line (B) Simple harmonic motion (C) Any motion at constant speed (D) Any single object in motion with constant momentum

13. A rope of negligible mass supports a block that weighs 30 N, as shown above. The breaking strength of the rope is 50 N. The largest acceleration that can be given to the block by pulling up on it with the rope without breaking the rope is most nearly (A) 6.7 m/s2 (B) 10 m/s2 (C) 16.7 m/s2 (D) 26.7 m/s2

Questions 14-15 A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is hanging from the board. The tension in the right chain is 250 N.

14. What is the tension in the left chain? (A) 125 N (B) 250 N (C) 375 N (D) 625 N

41 15. Which of the following describes where the person is hanging? (A) between the chains, but closer to the left-hand chain (B) between the chains, but closer to the right-hand chain (C) Right in the middle of the board (D) directly below one of the chains

16. Multiple correct: The cart of mass 10 kg shown above moves without frictional loss on a level table. A 10 N force pulls on the cart horizontally to the right. At the same time, a 30 N force at an angle of 60° above the horizontal pulls on the cart to the left. Which of the following describes a manner in which this cart could be moving? Select two answers.

(A) moving left and speeding up (B) moving left and slowing down (C) moving right and speeding up (D) moving right and slowing down

17. Two people are pulling on the ends of a rope. Each person pulls with a force of 100 N. The tension in the rope is: (A) 0 N (B) 50 N (C) 100 N (D) 200 N

18. The parabola above is a graph of speed v as a function of time t for an object. Which of the following graphs best represents the magnitude F of the net force exerted on the object as a function of time t?

(A) (B) (C) (D)

1976 B1 (modified) An elevator car that has a mass of 400 kg is in contact with two rails that each exert a frictional force of 200 N on the car. The elevator is attached to a cable that passes over a pulley with negligible friction and is attached to a counterweight M. Under these m conditions, the elevator has an upward acceleration of 2 2 . s a. On the diagrams below draw all the forces acting on the elevator car and the counterweight. Identify the source of all the forces.

b. Write two equations that relate your free body diagrams to the mass of the elevator car m, the mass of the counterweight M, and the acceleration a, of the system.

Elevator (forces directed up positive) ΣF = ma

FT − 2Ff − mg = ma

Counterweight (forces directed down positive) ΣF = Ma Mg − F = Ma T

c. Calculate the tension in the cable supporting the elevator.

FT − 2Ff − mg = ma

FT = 2Ff + mg + ma ⎛ m⎞ ⎛ m⎞ F = 2 200N + 400kg 10 + 400kg 2 T ( ) ⎝⎜ s2 ⎠⎟ ⎝⎜ s2 ⎠⎟ F = 5200 N T

d. Calculate the mass M, of the counterweight.

Mg − FT = Ma

Mg − Ma = FT M g − a = F ( ) T F 5200 N M = T = = 650 kg g − a ⎛ m m⎞ ( ) 10 2 ⎜ 2 − 2 ⎟ ⎝ s s ⎠

2007B (modified) A child pulls a sled along the ground with a force of 200 N which makes an angle θ = 37° with the horizontal. A dog rides on top of the sled so that the dog-sled system has a total mass of 40 kg. The coefficient of friction between the sled and the ground is 0.5. Use sin(37°) = 0.6 and cos(37°) = 0.8.

a. On the diagram below, draw the forces acting on the dog-sled system. Identify the source of all the forces.

b. Write two equations that relate your free body diagram to the mass of the dog-sled system m, and the acceleration a, of the dog-sled system. Acceleration of dog-sled system is horizontal Vertical Forces: ΣF = 0

FN + FT sinθ − mg = 0

FN + FT sinθ = mg Horizontal Forces: ΣF = ma F cosθ − F = ma T f

c. Qualitatively discuss what happens to the normal force as the angle, θ is increased or decreased. Calculate the normal force acting on the dog-sled system.

As the angle, θ increases the vertical component of the tension force increases. Hence, the normal force decreases. As the angle, θ decreases the vertical component of the tension force decreases. Hence, the normal force increases.

FN + FT sinθ = mg

FN = mg − FT sinθ ⎛ m⎞ F 40kg 10 200N sin37 280 N N = ( )⎜ 2 ⎟ − ( )( °) = ⎝ s ⎠

d. Calculate the friction force acting on the dog-sled system.

F = µF = 0.5 280N = 140 N f N ( )

e. Calculate the acceleration of the dog-sled system.

FT cosθ − Ff 200N cos37° −140N m a = = = 0.5 2 m 40kg s