Items Covered

Date Item

14

15

16

9/11 Lab 3 — Determine the initial speed of the rocket

9/16Lab 4 — Predict the landing spot of a projectile that is

launched horizontally

10/1Lab 5 — Predict the landing spot of a projectile that is

launched at an angle.

AP

PHYS

Items Covered

Date Item

17

18

19

10/07 The Forces of Nature

10/08 Calculating the Normal Force

10/13 Lab 7 — Determining the Inertial Mass of a Loaded Cart.

AP

PHYS

Items Covered

Date Item

20

21

22

10/22/15 Elevator Man

10/26/15 Friction

10/26/15 Lab 8 — Determining the Coefficient of Friction

AP

PHYS

PHYSICS

Items Covered

Date Item

23

24

25

11/02 4.8 — Law of Universal Gravitation

11/05/15 Atwood Machine

11/06/15 Lab 9 — Atwood’s Machine

AP

PHYS

= 47 N = 77N − 30N a) FN = m1g − FT

Fg + FT + FN = m1a1

A box weighing 77.0 N rests on a table. A rope tied to the box runs vertically upward over a pulley and a weight is hung from the other end. Determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weight (a) 30.0 N, (b) 60.0 N, and (c) 90.0 N.

AP

PHYS

Warm Up (11/9/15) — Atwood’s Machine

0, from rest!

27. (II) A train locomotive is pulling two cars of the same massbehind it, Fig. 4–51. Determine the ratio of the tension inthe coupling (think of it as a cord) between the locomotiveand the first car to that between the first car and thesecond car for any nonzero acceleration of the train.AFT2B,AFT1B,

10.0°

FIGURE 4;47 Problem 23.

T2FB

T1FBCar 2 Car 1

FIGURE 4;51 Problem 27.

90°

x

y

120°

x

y

(a) (b)

1

2

1

2

FB

FB

FB

FB

FIGURE 4;52 Problem 28.

21. (I) Draw the free-bodydiagram for a basketballplayer (a) just before leav-ing the ground on a jump,and (b) while in the air.See Fig. 4–46.

102 CHAPTER 4 Dynamics: Newton’s Laws of Motion

26. (II) Two snowcats in Antarctica are towing a housing unitnorth, as shown in Fig. 4–50. The sum of the forces

and exertedon the unit by thehorizontal cablesis north, parallel to the line L,andDetermine andthe magnitude ofFB

A + FB

B .

FB

FA = 4500 N.

FB

BFB

A

22. (I) Sketch the free-body diagram of a baseball (a) at themoment it is hit by the bat, and again (b) after it has leftthe bat and is flying toward the outfield. Ignore air resistance.

23. (II) Arlene is to walk across a “high wire” strung horizontallybetween two buildings 10.0 m apart. The sag in the ropewhen she is at the midpoint is 10.0°, as shown in Fig. 4–47.If her mass is 50.0 kg, what is the tension in the rope at thispoint?

28. (II) The two forces and shown in Fig. 4–52a and b(looking down) act on an 18.5-kg object on a frictionlesstabletop. If and find the netforce on the object and its acceleration for (a) and (b).

F2 = 16.0 N,F1 = 10.2 N

FB

2FB

1

FIGURE 4;46Problem 21.

24. (II) A window washer pulls herself upwardusing the bucket–pulley apparatus shown inFig. 4–48. (a) How hard must she pull down-ward to raise herself slowly at constantspeed? (b) If she increases this force by 15%,what will her acceleration be? The mass ofthe person plus the bucket is 72 kg.

25. (II) One 3.2-kg paint bucket is hanging by a massless cordfrom another 3.2-kg paint bucket, also hanging by a mass-less cord, as shown in Fig. 4–49. (a) If thebuckets are at rest, what is the tension in each cord? (b) If the two buckets are pulled upward with an acceleration of by the upper cord, calculate the tension in eachcord.

1.25 m!s2

FIGURE 4;49Problem 25.

FIGURE 4;48Problem 24.

32°48°

L

Top view

A

B

FB

FB

FIGURE 4;50Problem 26.

4;7 Newton’s Laws and Vectors [Ignore friction.]20. (I) A box weighing 77.0 N rests on

a table. A rope tied to the box runsvertically upward over a pulley anda weight is hung from the otherend (Fig. 4–45). Determine theforce that the table exerts on the box if the weight hanging on the other side of the pulleyweighs (a) 30.0 N, (b) 60.0 N, and(c) 90.0 N.

FIGURE 4;45Problem 20.

Fg

FTFN = ?

Fnet = ma; system: m1

-m1g + FT + FN = 0

FN = m1g − FT m2g m1

m2

b) FN = 77N − 60N = 17 N

c) FN = 0

Fnetext

= ma ; system: mA

FgA + FTA = mAaA

+ FTA= +mAaA mB > mA

Fnetext

= ma ; system: mB

FgB + FTB = mBaB

+ FTB = -mBaB

mAaA = FTA − g

-mBg

mBaB = g − FTB

Atwood’s Machine

-mAg

Since aA = aB = a, then

mA FTA − g

mB= g − FTB

FT ( 1/mA + 1/mB ) = 2gNote that : FTA = FTB = FT FT = 2g

( 1/mA + 1/mB )

+( mB + mA )

FT = 2g mAmB

a =g −( mB + mA ) 2g mA

AP

PHYS

Lab 9 — Atwood’s MachineExperiment 5: Atwood’s Machine

In 1784, George Atwood created a device to calculate force and tension and to verify the laws of motionof objects under constant acceleration. His device, now known as an Atwood’s Machine, consisted of twomasses, m1 and m2, connected by a tight string that passes over a pulley, as seen in Figure 1. When themasses are equal, the pulley system is in equilibrium, i.e. balanced. When the masses are not equal, bothmasses will experience an acceleration (indicated with red arrows in Figure 1). For simplification, we willassume the ideal pulley scenario, where the mass of the string is negligible and we ignore any frictionaleffects acting on the pulley. With these assumptions, the accelerations a1 and a2 are equal.

m1m2

a2

a1

Figure 1: An Atwood Machine.

When m1 > m2, the sign convention used is such that m1 accelerates in the downwards y directionand m2 moves upwards. Here we let the acceleration be positive. On the other hand, if m1 < m2, we willset the acceleration as negative. With this convention, we can derive a system of equations describing theacceleration of each mass by applying Newton’s second law, F = ma, to each mass individually. Lookingat the free body diagrams for the masses of the Atwood machine (see Figure 2), we can write the forces as

m1a = m1g − T (1)

m2a = T −m2g (2)

where T is the tension in the string and g is the acceleration due to gravity (g = 9.8 m/s2).

1

a = g −( mB + mA ) 2g mA =

( mB + mA )g( mB − mA ) = Fnet

M

The objective of the lab is to study Newton’s Second Law (NSL) by confirming the result of the acceleration of the of the masses in the atwood’s machine predicted by NSL.

AP

PHYS

Lab 8 — Atwood’s Machine

ap =( mB + mA )

g( mB − mA )

m1 (kg) m2 (kg) ap (m/s2) am (m/s2) Time (s) % Error

= 2⋅yam

t2

c pulls on the pulley_ These forces are equal in magnitude for a fric-tionless pulley, and their combined pull is to the right. This force is balanced by the force f\egOOIlUlley of the patient 's leg pulling to the left. The traction force F pulley on leg fanns an action/reaction pair w ith it\cgon pulley. so 50 N of traction means that f\ cg on pulley also has a

magnitude of 50 N.

SOLVE Two important properties af fopes, give n in Tactics Box 5.2, are that (I ) the tension equals the magnilUde of the Force pulling on its end and (2) the tension is the same throughollt the rope. Thus, if a hangi ng mass 111 pulls on the rope with its weight mg, the tension along the entire rope is T = mg. For a 4.2 kg hang ing mass, the tension is then T = mg = 4 1 N.

The pulley, in equilibrium, must satisfy Newton 's second law for the case where a = O. Thus

EXAMPLE 5 19 Lifting a stage set A 200 kg set used in a play is stored in the loft above the stage. The rope holding the set passes up and over a pulley, then is tied backstage. The director tell s a 100 kg stagehand to lower the set. When he unti es the rope , the set fall s and the unfortunate man is hoi sted into the loft. What is the stagehand 's accelerat ion?

PREPARE FIGURE 5.38 shows the visual overview. The objects of interest are the stagehand M and the set S, for which we 've drawn separate free-body diagrams. Assume a mass less rope and a massless, frictionless pulley. Tension forces Ts and TM are due to a massless rope go ing over an ideal pulley, so their magnitudes are the same.

FIGURE 5.38 Visual overview for the stagehand and set.

Known 111 M - 100 kg IllS = 200 kg

Find

y

M

'--------,.

Since the rope is massless and (he pulley ideal. the magnitudes of these two tensions are the same. , 1\' i,

TM '/

M s

SOLVE From the two free-body diagrams, we can write Newton's second law in componen t form. For the man we have

5.8 Ropes and Pulleys 155

The tension forces both have the same magnitude T, and both are at angle 0 from horizontal. The x-component of the leg force is nega-tive because it 's directed to the left. Then Newton 's law becomes

so that

2TcosO - FlcgonPUllcy = 0

F 1eg on pulley cosO = 2T

SON 82 N

ASSESS The traction force would approach 2mg = 82 N if angle o approached zero because the two tensions would pull in paral-le l. Converse ly, the traction force would approach 0 N if 0 approached 90°. Because the desired traction force is roughly halfway between 0 Nand 82 N, an angle near 45° is reasonable.

For the set we have

Only the y-equations are needed. Because the stage hand and the set are connected by a rope, the upward di stance traveled by one is the same as the downward distance traveled by the other. Thus the maglliludes of their acceleratio ns must be the same, but , as Figure 5.38 shows, their directions are opposi te. We can express lhi s mathematically as asy = -OM.,'. We al so know that the two tens ion forces have equal magnitudes, which we' ll call T. Insert-ing [his informat ion into the above equatio ns gives

T - 11/Mg = 11/ M ClMy

T - "' sg = -mSaMy

These are simultaneous equations in the two unknowns T and aMy- We can solve for T in the first equation to get

Inserti ng this value of T into the second equation then gives

which we can rewrite as

FinaJly, we can so lve for the hapless stagehand's acceleration:

IllS- 111 M (IOOkg) , , aMI' = g = --- X 9.80 m/s- = 3.3 m/s -

111 S + 111 M 300 kg

This is also the acceleration with which the set falls. If the rope 's tension was needed, we could now find it from T = m MClMy + mMg.

ASSESS If the stagehand weren ' t ho lding on, the set would fall with acceleration g. The stagehand acts as a cOllllfer-weight to reduce the acceleration.

A 200 kg set used in a play is stored in the loft above the state. The rope holding the set passes up and over a pulley, then is tied backstage. The director tells a 100 kg stagehand to lower the set. When he unties the rope, the set falls and the unfortunate man is hoisted into the loft. What is the stagehand’s acceleration?

AP

PHYS

Warm Up (11/10/15) — Atwood’s Machine

Fg

FT

Starting from the equation that we derived last Thursday, we get:

ap =( mB + mA )

g( mB − mA )

(200kg + 100kg)

= 9.8m/s2( 200kg − 100kg) = 3.3 m/s2