Name________________________________________
Physics 120 Quiz #4, May 2, 2014
Please show all work, thoughts and/or reasoning in order to receive partial credit. The quiz is
worth 10 points total.
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____________________________
A mass of 105kg is suspended from a vertically oriented tungsten wire with dimensions
1mm × 2mm × 0.8m and the long axis of the tungsten wire is the vertical direction. The mass stretches
the piece of tungsten wire by 1mm in the vertical direction
g
kg
a. If tungsten has a molar mass of 184 mol and a density of 19250 m 3 , what is the diameter of an
atom of tungsten?
The diameter of the tungsten atom is calculated from the density.
ρ=
m
m
→ V = d3 = =
V
ρ
1mol
6.02 × 10 23 atom → d = 2.51× 10 −10 m .
19250 mkg3
kg
0.184 mol
×
b. What is the overall stiffness of the tungsten wire?
The stiffness is calculated by applying the momentum-principle to the situation. Here we have a
static situation and taking the positive y-direction pointing along the vertical axis of the wire we
!
dp !
mg 105kg × 9.8 sm2
= Fnet → 0,0,0 = 0, ky − mg,0 → kwire =
=
= 1.03 × 10 6
have,
dt
y
1× 10 −3 m
N
m
.
c. What is the stiffness of an interatomic bond in the tungsten wire?
To calculate the stiffness of an interatomic bond in tungsten, we need to know how many side-by-
A 1× 10 −3 m × 2 × 10 −3 m
= 3.17 × 1013 . We
side chains of atoms we have. We have N ' = 2 =
2
−10
d
( 2.51× 10 m )
also need to know the number of interatomic bonds in a single strand of atoms. We find
L
0.8m
=
= 3.19 × 10 9 . The stiffness of the interatomic bond comes from
−10
d 2.51× 10 m
1
N
examining a chain of springs in series,
, where the stiffness of a chain of atoms is
=
kchain kIAB,W
N=
related to the number of side-by-side chains we have of springs in parallel, kwire = N ' kchain .
Putting the last two results together we have
⎛ 3.19 × 10 9 ⎞
⎛N⎞
kIAB,W = ⎜ ⎟ kwire = ⎜
× 1.03 × 10 6
13 ⎟
⎝ N '⎠
⎝ 3.17 × 10 ⎠
N
m
= 103.7 Nm .
d. What is Young’s modulus for tungsten?
Young’s modulus is given by Y =
k IAB
103.6 Nm
=
= 4.13 × 1011 mN2 .
−10
d
2.51× 10 m
Physics 120 Equation Sheet
!
!
r =< rx ,ry ,rz >= r ⋅ r̂
!
magnitude of a vector : r = r = rx2 + ry2 + rz2
!
r
unit vector : r̂ = !
r
! !
!
v + vf
! Δr !
v=
; vavg = i
Δt
2
! ! !
rf = ri + vavg Δt
!
!
Δp
Fnet =
Δt
!
! !
p f = pi + Fnet Δt
!
Fnet
! ! !
rf = ri + vi Δt +
( Δt )2
2m
!
GM 1 M 2
FG = −
r̂
r2
!
Fg ~ mg!
Constants:
g = 9.8 sm2
G = 6.67 × 10 −11 Nm
kg 2
2
me = 9.11× 10 −31 kg
m p = 1.67 × 10 −27 kg
mE = 6 × 10 24 kg
RE = 6.4 × 10 6 m
N A = 6.02 × 10 23
!
dp !
= Fnet
dt
!
!
Fs = −ks
Ffr = µ FN
N
k parallel = ∑ ki
i=1
1
N
=∑
kseries i=1
m
ρ=
V
k
Y = IAB
d
1
ki