Homework: 3, 5, 7, 13, 24, 34, 45, 49, 51, 56, 57, 59

3. If the 1 kg standard body has an acceleration of 2.00 m/s2 at 20.00 to the positive direction of an x axis, what are (a) the x component and (b) the y component of the net force acting on the body, and (c) what is the net force in unit-vector notation?

acosθ ma mF xx

asinθ ma mF yy

)jsinθia(cosθ mjFiFF yxnet

5. Three astronauts propelled by jet backpacks, push and guide a 120 kg asteroid toward a processing dock, exerting the forces shown in the figure below, with F1=32 N, F2=55 N, F3=41 N, 1=30

0, and 3=600. What is the asteroid’s acceleration (a) in unit-

vector notation and as (b) a magnitude and (c) a direction relative to the positive direction of the x axis?

xx a mF

33112x cosθFcosθFFF

3311y sinθFsinθFF

yy a mF

jaiaa yx

2y

2x aaa

x

y1

a

atanθ

13. The figure below shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 98 N on the wall to which it is attached. The tension in the shorter cords are T1=58.8 N, T2=49.0 N, and T3=9.8 N. What are the masses of (a) disk A, (b) disk B, (c) disk C, and (d) disk D?

T

Disk A: T=T1+mAg mA=4.0 (kg)

Disk B: T1=T2+mBg mB=1.0 (kg)

Disk C: T2=T3+mCg mC=4.0 (kg)

Disk D: T3=mDg mD=1.0 (kg)

24. There are two horizontal forces on the 2.0 kg box in the overhead view of the figure below but only one (of magnitude F1=30 N) is shown. The box moves along the x axis. For each of the following values for the acceleration ax of the box, find the second force in unit-vector notation: (a) 10 m/s2, (b) 20 m/s2, (c) 0, (d) -10 m/s2, and (e) -20 m/s2.

F1+F2=ma

(a) F2=ma-F1=2.0x10-30=-10 (N)

i)N10(F2

(d) F2=ma-F1=2.0x(-10)-30=-50 (N)

i)N50(F2

34. In the figure below, a crate of mass m=115 kg is pushed at constant speed up a frictionless ramp (=30.00) by a horizontal force . What are the magnitudes of (a) and (b) the force on the crate from the ramp?

F

F

Fg

FN

(a) The crate moves with a constant speed, so the net force acting on the crate is zero. Along the x axis: Fcos - mgsin = ma + a = 0 (constant speed): Fcos = mgsin F = 651 (N)

(b) Along the y axis: FN - mgcos - Fsin = 0 FN = mgcos + Fsin FN = 1302 (N)

x

y

45. An elevator cab that weighs 27.8 kN moves upward. What is the tension in the cable if the cab’s speed is (a) increasing at a rate of 1.22 m/s2 and (b) decreasing at a rate of 1.22 m/s2?

(a) Applying Newton’s second law, a=+1.22 m/s2: T-mg = ma m = (27.8x1000)/9.8 = 2837 (kg) T = 2837 (9.8+1.22) = 31.3 x 103 (N) (b) a=-1.22 m/s2: T = 2837 (9.8-1.22) = 24.3 x 103 (N)

+

51. The figure below shows two blocks connected by a cord (of negligible mass) that passes over a frictionless pulley (also of negligible mass). The arrangement is known as Atwood’s machine. One block has mass m1=1.3 kg; the other has mass m2=2.8 kg. What are (a) the magnitude of the block’s acceleration and (b) the tension in the cord?

F1,g

T

F2,g

T

y

m1g - T = m1a1

m2g – T = m2a2

)m/s(6.3mm

)gm(ma 2

21

12

T = m1(g+a) T = 17.4 (N)

a1 = -a2 = -a: m1g - T = -m1a

m2g – T = m2a

56. In Figure a, a constant horizontal force is applied to block A, which pushes against block B with a 15.0 N force directed horizontally to the right. In Figure b, the same force is applied to block B; now block A pushes on block B with a 10.0 N force directed horizontally to the left. The blocks have a combined mass of 12.0 kg. What are the magnitudes of (a) their acceleration in Figure a and (b) force ?

(a) Figure a: FB= mBa Figure b: F’A= mAa a = (F’A+FB)/(mA+mB) FB=15 N; F’A=10 N; mA+mB = 12 kg a = 2.08 (m/s2) (b) Fa = (mA+mB)a = 25 (N)

10 N FB=15 N

Additional question: what are masses mA and mB?

mA = F’A/a = 10/2.08 4.8 kg

mB = FB/a = 15/2.08 7.2 kg

F’A=10 N

Inverse problem:

If we know Fa = 25 N, mA = 4.8 kg and mB = 7.2 kg, Determine contact forces between the blocks in Figure a and b.

Fa = (mA+mB)a a = Fa/(mA+mB) = 25/12 2.08 m/s2

Figure a: FB = mBa FB= 7.2 x 2.08 15 N

Fa-FA = mAa FA= 25 – 4.8 x 2.08 15 N

Figure b: F’A = mAa F’A= 4.8 x 2.08 10 N

Fa-F’B = mBa F’B= 25 – 7.2 x 2.08 10 N

F’A FB FA F’B

57. A block of mass m1=3.7 kg on a frictionless plane inclined at angle =30.00 is connected by a cord over a massless, frictionless pulley to a second block of mass m2=2.30 kg hanging vertically. What are (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord?

1. Force analysis

NF

g1,F

T

'T

g2,F

y x

y

2. Applying Newton’s second law:

0θcosFF g1,N

amθsinFT 1g1,

Block 1:

Block 2: amT-F 2g2,

)(735.0 2

21

12 m/smm

gsinθm-gma

a>0: the direction of the acceleration of block 2 is downward.

(N)20.9a)-(gmam-FT 22g2,

59. A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground (a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground? If, after the package has been lifted, the monkey stops its climb and holds onto the rope, what are the (b) magnitude and (c) direction of the monkey’s acceleration and (d) the tension in the rope?

Fm

T

T FN

Fp

(a) T: the force the rope pulls upward on the monkey: T – mg = mam

For the package: T+FN-Mg = Map

To lift the package off the ground: FN=0, and the least acceleration am requires ap=0, so: T = Mg Mg – mg = mam

am = 4.9 (m/s2) (b) See Problem 51: (c) See Problem 51: T = m(g+am) 118 (N)

)m/s(96.1mM

m)g(Ma 2

Chapter 2 Force and Motion

2.1. Newton’s First Law and Inertial Frames 2.2. Newton’s Second Law 2.3. Some Particular Forces. The Gravitational Force and Weight 2.4. Newton’s Third Law

2.5. Friction and Properties of Friction. Motion in the Presence of Resistive Forces 2.6. Uniform Circular Motion and Non-uniform Circular Motion 2.5. Motion in Accelerated Frames

2.5. Friction and Properties of Friction. Motion in the Presence of Resistive Forces

• No motion of the block: : static frictional force

• Motion of the block: : kinetic frictional force

maxs,k ff

• Friction:

sf

kf sf

F

No motion

kf

'F

motion

• Properties of friction:

Property 1: If the body does not move, and the component of that is parallel to the surface are equal in magnitude and opposite in direction.

sf

F

Property 2: The magnitude of has a maximum value computed by: is the coefficient of static friction. is the magnitude of the normal force on the body from the surface.

sf

Nsmaxs, Fμf

sμ

NF

Ffs

Ffs

maxs,

s

f

Ff

0fs

• Properties of friction:

Property 3: If the body moves, the magnitude of the frictional force decreases to a value fk calculated by: is the coefficient of kinetic friction

Nkk Fμf

kμ

Nkk Ff

Nkk Ff

• Sample Problem: A woman pulls a loaded sled of m=75 kg at constant speed; k=0.10; =420; determine: (a) (b) T increases, how about ?

Nkkk Ff 0;f-TcosΦ

Constant speed requires a = 0, so: • For the x axis:

T

kf

• For the y axis:

(2)0mgFTsinΦ N

(N)7.90sinΦμcosΦ

mgμT(2) & (1)

k

k

TsinΦ-mgFN

If T increases, FN will decrease fk decreases

loaded sled

amFnet

)1(0F-TcosΦ Nk

• Checkpoint: F1=10 N, F2 increases from 0. Before the box begins to slide, do the following quantities increase, decrease or stay the same: (a) fs; (b) FN; (c) fs,max

(a) the same; (b) FN+F2=mg FN decreases; (c) , so fs.max decreases Nsmaxs, Fμf

2F

Summary Steps for solving problems using Newton’s laws

1. Draw a free-body diagram for each object of the system: - draw all possible forces: gravitational, normal, tension, friction

(static or kinetic), any applied forces, third-law force pairs. - choose a coordinate system for each moveable object. - indicate the acceleration direction of each object, if unknown you

can make an assumption. 2. Write Newton’s second law: - Write the equation above for each axis - If the system is stationary or moving with a constant speed, then

a = 0. 3. Put constraints on the accelerations of the objects

amFnet

;; ,, yynetxxnet maFmaF

• Motion in the Presence of Resistive Forces:

If a body moves through a fluid (gas or liquid), the body will experience a drag force (due to air or viscous resistance) that opposes the relative motion.

D

• Drag at high velocity: is the density of the fluid v is the speed of the body relative to the fluid A is the effective cross-sectional area C is the drag coefficient

2AvCρ2

1D

- For a body falling through air: Fg’ – D = ma Fg’ = mg - Fbuoyant

D ~ v2

D increases until D=Fg’, and the body falls at a constant speed, called the terminal speed Vt:

ACρ

Fv

AvCρ2

1F

g't

2tg'

2

0

+

• Drag at low velocity: b is a constant, depending on the properties of the fluid and the dimension of the body v is the speed of the body

bvD

(1) mabv-F-mg buoyant

• D increases until the acceleration a=0: )2(bvF-mg tbuoyant

dt

dvmv)-b(vor mav)-b(v (2) and (1) tt

t

0

v

0 tt

dtm

b

v-v

dvdt

m

b

v-v

dv

)3()e(1vvtm

b

v

v-vln

tm

b

t

t

t

);e(1b

mg'v

tm

b

b

mg'

b

Fmg v (1)

buoyant

t

tm

b

eg'a

1)-(eb

mvtvy

tm

b

tt

(3)

1)-τ(evtvy τ

t

tt

timesticcharacteri the:b

mτ

)e(1vv τ

t

t

τ

t

eg'a

Homework: Read Sample Problem (p 123) 5, 9, 19, 25, 31, 34, 39 (p 130-134)