Physics 203 – Section B5
QUIZ 3
9 June 2014
Name:
1. (4 points) A person is trying to judge whether a picture (mass m = 1.10 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the
wall is 0.66. What is the minimum amount of pressing force that must be used?
Solution: The sum of the forces are
∑ Fx
∑ Fy
= F − FN = ma x = 0
(1)
= Ffric − mg = may = 0,
(2)
where F is the force with which you are pressing the picture against the wall
and both accelerations are zero because the picture is stationary. The picture is
stationary, so we are dealing with static friction. For static friction, Ffric ≤ µs FN .
From (1), the normal force FN = F, so Ffric ≤ µs F. For minimum pressing force
F, the force of friction Ffric = µs F exactly. Then substituting this into (2),
Ffric = µs F = mg =⇒ F =
(1.10 kg)(9.8 m/s2 )
mg
=
= 16.3333 N.
µs
0.66
(3)
2. (4 points) Two particles are located a distance 1.00 m apart. The first particle has
mass m1 = 1.50 kg and the second particle has mass m2 = 3.00 kg. A third particle
of mass m3 = 1.00 kg is placed in between the first and second particle (all three
lie on the same line). How far away from the first particle of mass m1 should the
third particle of mass m3 be placed such that the net gravitational force on the third
particle is zero?
Solution: The force from the first particle on the third particle
F31 =
Gm3 m1
Gm3 m1
=
2
x2
r31
(4)
pointing from the third particle to the first particle, where x is the distance from
between the first and third particles. The force from the second particle on the
Physics 203 – Section B5
QUIZ 3
9 June 2014
third particle
F32 =
Gm3 m2
Gm3 m2
=
2
( L − x )2
r32
(5)
pointing from the third particle to the second particle, where L = 1.00 m between
the first and second particle. These are the only two forces acting on particle
three. The net force on particle three, which we are told is zero is
Fnet = F32 − F31 =
Gm3 m2
Gm3 m1
−
= 0.
2
( L − x)
x2
(6)
The negative sign comes from one force being in one direction and the other
being in the opposite direction (remember, force is a vector, so you must always
take into account the direction, even if two forces are in one dimension).
The equation (6) is equivalent to
Gm3 m2
Gm3 m1
=
.
2
( L − x)
x2
(7)
Multiplying both sides by x2 ( L − x )2 and then taking the square root of both
sides gives the linear equation
x
p
Gm3 m2 = ( L − x )
p
Gm3 m1 .
(8)
Now we can solve for the distance between the first and and third particle
p
√
√
1.00 m 1.50 kg
L m1
L Gm3 m1
p
√
x= √
= 0.414214 m.
=√
=p
√
m2 + m1
Gm3 m2 + Gm3 m1
3.00 kg + 1.50 kg
(9)
3. (4 points) A jetliner flying at 125 m/s banks at an angle of 10◦ from the horizontal
to make a horizontal circular turn. The radius of the turn is 3610 m and the mass of
the jetliner is 2.00 × 105 kg. Calculate the magnitude of the necessary lifting force.
(Hint: this is the same problem as a car going around a bank but with the normal
force from the ground replaced by the force of lift from the air.)
Page 2
Physics 203 – Section B5
QUIZ 3
9 June 2014
Solution: First of all, we know that the acceleration in the vertical direction is
0 m/s2 and the acceleration in the horizontal direction is ac . Therefore, I want to
set my coordinate system with a x in the horizontal direction and ay in the vertical
direction. There are two forces acting on the plane. First is the plane’s weight m~g
pointing downward. Next there is the force of lift ~Flift that points “up” from the
plane. In this case, the plane is banked at an angle θ = 10◦ , so ~Flift is pointing 10◦
from the vertical. The sum of the forces
mv2
r
= Flift cos θ − mg = may = 0.
∑ Fx = Flift sin θ = max = mac =
(10)
∑ Fy
(11)
I made a mistake in the statement of the problem by giving you an extra, inconsistent piece of information. Therefore, there are two solutions for Flift depending
on how you proceed (both results were acceptable answers).
Solving (11) gives
Flift =
mg
(2.00 × 105 kg)(9.8 m/s2 )
=
= 1.990 24 × 106 N.
cos θ
cos 10◦
(12)
Solving (10) instead gives
Flift =
(2.00 × 105 kg)(125 m/s)2
mv2
=
= 4.985 09 × 106 N.
◦
r sin θ
(3610 m) sin 10
Page 3
(13)