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 = M1 g (1)

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

Ff = µM1 g . (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
F − Ft
F1 = M1 a = F − Ff =⇒ a=
M1
Ft
F2 = M2 2 = Ff =⇒ a= (5)
M2
so if we eliminate the acceleration from these formulas we get
F − Ft Ft M1 + M2 M1 + M2
= =⇒ F = Ff = µM1 g (6)
M1 M2 M2 M2

(b) If the force is applied to the bottom block, then the bottom block feels the total
force
F2 = F − Ff = M2 a (7)

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 while the top block feels the total force

F1 = f = M1 a . (8)

Again, both blocks must have the same acceleration so the force F equals
M1 + M2 M1 + M2
F = Ff = µM1 g = (M1 + M2 )µg (9)
M1 M1

2. Problem 2.9

N
θ
F mg

Since the particle is moving on a circle in the horizontal plane, vertical component of
the normal force N must cancel the gravitational force Fg = mg so that the total force
is always in the horizontal plane

N sin θ = mg . (10)

Since the remaining force is centripetal

mv02
F = = N cos θ (11)
r
we get the radius
mv02 v02
r= = tan θ . (12)
N cos θ g
3. Problem 2.11

(a)

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 Tup

45o

Ftot o
45
θ mg
Tlow

(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 θ − mg − Tlow cos θ (13)

while the sum of horizontal components must equal centripetal force

Tup sin θ + Tlow sin θ = mω 2 l cos θ . (14)

√
For θ = 45◦ , cos θ = sin θ = 1/ 2 so

mω 2 l mg mω 2 l mg
Tup = +√ , Tlow = −√ (15)
2 2 2 2
√
For Tlow > 0, we must have ω 2 > 2g/l.

4. Problem 2.13

l1
l’1

l’2
x1
l2
x2

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 Let’s denote the tension of the string on the first pulley T and on the second T 0 .
Acceleration of the mass M1 is then given by

ẍ1 M1 = T − M1 g (16)

while the acceleration of the second one is given by

ẍ2 M2 = T 0 − M2 g . (17)

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

T = 2T 0 . (18)

Finally, since both strings are inextensible,

l1 + l10 = L1 , , l2 + l20 = L2 (19)

sum of lengths
(x1 + l1 + l10 ) + (l20 + x2 ) = const. (20)
| {z }
= 21 (l2 +l20 +x2 )

must be constant, we have

ẍ2
ẍ1 = − (21)
2
so now we can eliminate the string tension T
6M1 M2
T = g (22)
M1 + 4M2
and then evaluate the acceleration ẍ1 to get
2M2 − M1
ẍ1 = g. (23)
4M2 + M1

5. Problem 2.14
Let’s call the string tension connecting masses MA and MB T , and the string connecting
mass MC with the pulley T 0 . Then masses MA and MB feel only force T

T = MA ẍA , T = MB ẍB . (24)

Third mass MC feels the effective force mg − T 0

mg − T 0 = Mc ẍc . (25)

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 xB xp

xA MA MB
xC

MC

Since both pulleys are massless, we must have T 0 = 2T . Finally, since strings are
inextensible, we have xc − xp = LC = const., where
1
xp = xA + lA = xB + lB = (xA + xB + LAB ) (26)
2
where LAB is the length of first string and LC is the length of second. Hence

ẍA + ẍB = 2ẍC . (27)

From this we can find string tension T

2MA MB MC
T = (28)
4MA MB + MA MC + MB MC
from which we get accelerations
2MB MC
ẍA = g (29)
4MA MB + MA MC + MB MC
2MA MC
ẍB = g (30)
4MA MB + MA MC + MB MC
(MA + MB )MC
ẍC = g (31)
4MA MB + MA MC + MB MC

6. Problem 2.17

(a) Normal force to the surface is given by

N = mg cos θ (32)

which gives us the friction force

F < Fmax = µN = µmg cos θ . (33)

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 F N
F||

mg

On the other hand, force parallel to the surface equals

Fk = mg sin θ (34)

so for the block to rest we must have

Fk ≤ Fmax =⇒ tan θ ≤ µ (35)

(b) If we add the acceleration in the horizontal direction, then vertical forces ad up
to
F sin θ + N cos θ − mg = 0 (36)

while horizontal forces add up to

N sin θ − F cos θ = ma (37)

This can be solved for a

sin θ − µ cos θ
amin = g (38)
cos θ + µ sin θ
There is a typo in the textbook and the actual condition in part (b) of this
problem should read tan θ > µ for this to make sense. The way angles are drawn
in the textbook, this solution is negative if one takes tan θ < µ. 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 θ < µ 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.
(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

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 the sign of F we now have

0 = −F sin θ + N cos θ − mg (39)

ma = N sin θ + F cos θ (40)

which has the solution

sin θ + µ cos θ
amax = g (41)
cos θ − µ sin θ

7. Problem 2.19
For M3 to keep from rising or falling, all 3 masses must move with the same (horizontal)
acceleration and vertical forces must vanish. Mass M3 feels two vertical forces, M3 g
pointing down and T pointing up; they must cancel so

T = M3 g . (42)

Mass M2 feels only string tension T so it accelerates with acceleration a

T = M2 a . (43)

Eliminating the string tension T , we get the acceleration

M3
a= g. (44)
M2
Since the total system has a combined mass M = M1 +M2 +M3 and there is no relative
motion of 3 masses, force acting on mass M1 must produce the overall acceleration a
for the total system, so
M3
F = g(M1 + M2 + M3 ) (45)
M2
8. Problem 2.29

ω
a
2v0 ω
v0 ω 2 t

(a) In polar coordinates, acceleration s given by

~a = (r̈ − rθ̇2 )r̂ + (rθ̈ + 2ṙθ̇)θ̂ (46)

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 Since the car is moving with uniform velocity v0 along the line painted radially
on the platform, we have

r = v0 t , ṙ = v0 , r̈ = 0 . (47)

The platform is rotating with constant angular speed ω so

θ = ωt , θ̇ = ω , θ̈ = 0 . (48)

Adding these up, we get

~a = −v0 ω 2 tr̂ + 2v0 ω θ̂ (49)

(b) This part of the question can be a little confusing; friction coefficient µ describes
the fiction as tires of the car start sliding relative to the ground. Although the
car is moving there is no movement friction involved as long as tires are rolling
normally and not skidding (in reality, there is some friction associated with tires
rolling down the road, but we neglect it), so we treat friction force as static
friction.1 Car starts to skid when friction µN = µM g equals force the car feels
M |~a| or in another words
1 q
(v0 ω)2 (4 + ω 2 t2 ) = (µg)2 =⇒ t= (µg)2 − 4v02 ω 2 . (50)
v0 ω 2

F
α
r
(c) Since the friction car feels in part (b) is essentially static friction, force will have
the the same size and direction as the acceleration (if there was no friction what-
soever, the car would continue to move in the radial direction as the platform
turns). Angle of the acceleration vector is given by
aθ 2v0 ω 2 2v0 ω
tan α = = = = q (51)
ar v0 ω 2 t ωt (µg)2 − 4v02 ω 2
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If this isn’t all too clear, think of yourself pushing a (static) car sideways; unless you are extremely
strong, chances are you will not be able to move it because the friction blocks you. Now think of a car going
at some speed low enough for you to run next to it and push it sideways (for example 5mph). What happens
as you push the car sideways? Again, the car doesn’t move sideways since the friction is still there and it
blocks your push. So the friction you encounter is still the static friction (even though the car is moving!)
equal to your push but in the reverse direction, not the friction of a moving body equal to µN and pointing
opposite the velocity.

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9. Problem 2.33
Since the particle is restricted to move on the rotating rod, its angular component is
θ(t) = ωt 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

d2 r
m(r̈ − rω 2 ) = 0 =⇒ = rω 2 (53)
dt2
which is satisfied by
r(t) = Ae−ωt + Beωt (54)

for any A and B. Initial conditions at t = 0 become

r(0) = A + B , v(0) = ω(B − A) . (55)

To get the solution which decreases in time, we must have B = 0. For any other choice,
exponentially growing term Beωt will eventually become bigger then exponentially
decaying term Ae−ωt .

10. Problem 2.34

r
T m

(a) Since there is no force acting on the particle in the angular direction θ̂, angular
component of the acceleration must vanish
ω̇ ṙ dω dr
aθ = 0 = rθ̈ + 2ṙθ̇ = rω̇ + 2ṙω =⇒ = −2 =⇒ = −2
ω r ω r
(56)
where ω = θ̇. This can easily be integrated
Zω Z r
dω dr ω r r2
= −2 =⇒ log = −2 log = log 02 (57)
ω0
ω r
r ω0 r0 r
0

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 where at some time t = 0 we have r = r0 and ω = ω0 . Trajectory of the particle
is then given implicitly by
ω r02
= 2 . (58)
ω0 r
Since the string is being pulled with constant velocity v, 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

r02
ω = ω0 . (60)
(r0 − vt)2

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

−T = m(r̈ − rω 2 ) . (61)

Since the particle moves along the trajectory r = r0 − vt, we have r̈ = 0 and

r04 r04
T = mrω 2 = mω02 = mω 2
0 (62)
r3 (r0 − vt)3

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