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Physics I (8.012) Fall 2004

Problem Set # 2 solutions

Due 9/24/2004

1. Problem 2.7

The vertical force between blocks is always equal to

Fv M1g (1)

so the maximal relative friction force we can have between two blocks is equal to

Ff M1g (2)

(a) If the external force is applied to the top block, the effective force it feels is the (vector)

sum of external force and friction force

F1 F

Ff (3)

According to the third Newton law, the friction force on the surface connecting top and

bottom blocks has a counter-force of the same magnitude but in the opposite direction,

F2

Ff

(4)

Since there is no slipping between two blocks, they both must have the same speed and

the same acceleration, so

F1 M1a

F

Ff a

F

Ft

M1

F2 M22

Ff a

Ft

M2(5)

so if we eliminate the acceleration from these formulas we get

F

Ft

M1

Ft

M2

F M1

M2

M2Ff

M1

M2

M2M1g (6)

(b) If the force is applied to the bottom block, then the bottom block feels the total force

F2 F

Ff M2a (7)

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45o

45o

mg

T

T

up

low

totF

(b) Vector sum of forces acting on the particle must act in the horizontal plane (as shown

in the picture). That means that vertical components must cancel

Tup cos

mg

Tlow cos

(13)

while the sum of horizontal components must equal centripetal force

Tup sin

Tlow sin m2l cos (14)

For 45 , cos sin 1

2 so

Tup

m2l

2

mg

2

Tlow

m2l

2

mg

2(15)

ForTlow

0, we must have 2

2g

l.

4. Problem 2.13

ll

l

x

1

1

x1

2

2

2

l

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Lets denote the tension of the string on the first pulley Tand on the secondT . Acceleration

of the massM1is then given by

x1M1 T

M1g (16)

while the acceleration of the second one is given by

x2M2 T

M2g (17)

If pulleys are massless then string tensions on the second pulley must cancel

T 2T (18)

Finally, since both strings are inextensible,

l1

l 1 L1

l2

l 2 L2 (19)

sum of lengths

x1

l1

l 1

l 2

x2

1

2

l2 l 2 x2

const (20)

must be constant, we have

x1

x2

2 (21)

so now we can eliminate the string tensionT

T

6M1M2

M1

4M2g (22)

and then evaluate the acceleration x1to get

x1

2M2 M1

4M2

M1g

(23)

5. Problem 2.14

Lets call the string tension connecting massesMAand MBT, and the string connecting mass

MCwith the pulley T . Then massesMA and MBfeel only force T

T MA xA

T MB xB (24)

Third massMCfeels the effective force mg T

mg

T Mc xc (25)

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x

x

x

A

pxB

C

M M

M

A B

C

Since both pulleys are massless, we must have T 2T. Finally, since strings are inextensi-

ble, we havexc xp LC

const , where

xp xA

lA xB

lB

1

2 xA

xB

LAB (26)

whereLAB is the length of first string andLCis the length of second. Hence

xA

xB 2 xC (27)

From this we can find string tensionT

T MAMBMC

2MAMB

MAMC MBMC

4

(28)

from which we get accelerations

xA

MBMC

2MAMB

MAMC MBMC

2

g (29)

xB

MAMC

2MAMB

MAMC MBMC

2

g (30)

xC

1

2

MA

MB MC

2MAMB

MAMC MBMC

2

g (31)

6. Problem 2.17

(a) Normal force to the surface is given by

N mgcos (32)

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||F

mg

NF

which gives us the friction force

F

Fmax N mgcos

(33)

On the other hand, force parallel to the surface equals

F

mgsin (34)

so for the block to rest we must have

F

Fmax tan

(35)

(b) If we add the acceleration in the horizontal direction, then vertical forces ad up to

Fsin

Ncos mg 0 (36)

while horizontal forces add up to

Nsin Fcos ma (37)

This can be solved fora

amin

sin cos

cos

sing (38)

There is a typo in the textbook and the actual condition in part (b) of this problem

should read tan

for this to make sense. The way angles are drawn in the textbook,

this solution is negative if one takes tan . This means that the acceleration is

pointing in the opposite direction from the way it is drawn in the textbook. This of

course has to be, since for tan block will remain motionless with no acceleration

and the answer is amin 0. What we actually calculated is the maximum acceleration

in the negative direction before the block starts sliding down.

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(c) As we increase the acceleration, sum of tangential forces at one point changes the

direction and starts pointing upward so friction now points downward. Changing the

sign ofFwe now have

0 Fsin

Ncos mg (39)

ma Nsin

Fcos (40)

which has the solution

amax

sin

cos

cos

sing (41)

7. Problem 2.19

ForM3 to keep from rising or falling, all 3 masses must move with the same (horizontal)

acceleration and vertical forces must vanish. MassM3feels two vertical forces,M3g pointing

down andTpointing up; they must cancel so

T M3g (42)

MassM2feels only string tension Tso it accelerates with accelerationa

T M2a (43)

Eliminating the string tensionT, we get the acceleration

a M3M2

g (44)

Since the total system has a combined mass M M1

M2

M3 and there is no relative

motion of 3 masses, force acting on massM1must produce the overall acceleration a for the

total system, so

F M3

M2g M1

M2

M3 (45)

8. Problem 2.29

2v

0

0

v t2

a

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9. Problem 2.33

Since the particle is restricted to move on the rotating rod, its angular component is t

t

and there is no angular component to the acceleration. Its trajectory is then described by

m

r 0 (52)

since there are no forces acting on the particle an it is free to move along the rod. Radial

component of this equation is

m r

r2

0 d2r

dt2 r2 (53)

which is satisfied by

r t

Ae

t Bet (54)

for anyA and B. Initial conditions att

0 become

r 0

A

B

v 0

B

A

(55)

To get the solution which decreases in time, we must have B 0. For any other choice,

exponentially growing termBet will eventually become bigger then exponentially decaying

termAe

t.

10. Problem 2.34

m

v

T

r

(a) Since there is no force acting on the particle in the angular direction, angular compo-

nent of the acceleration must vanish

a 0 r

2r r

2r

2r

r

d

2dr

r

(56)where . This can easily be integrated

0

d

2

r

r0

dr

r log

0

2log r

r0

logr20

r2 (57)

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where at some timet 0 we haver r0and

0. Trajectory of the particle is then

given implicitly by

0

r20

r2 (58)

Since the string is being pulled with constant velocityv, if we assume that the string is

inextensible, then the radius of the particle changes as

r t

r0 vt (59)

so angular velocity is given by

0r20

r0 vt 2 (60)

(b) ForceT needed to pull the the string is related to the angular acceleration as

T m r r2

(61)

Since the particle moves along the trajectoryr r0 vt, we have r 0 and

T mr2 m20r40

r3 m20

r40

r0 vt 3 (62)

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