Al Azhar university Lama El Banna general physic

1. General Physics for Medical Sciences
Homework 3 solution
Instructor: Dr. Hassan Ashour
1. Assume that Young’s modulus for bone is 1.5 × 10 N/m and that a bone will fracture if
more than 1.50 × 10 N/m is exerted. (a) What is the maximum force that can be
exerted on the femur bone in the leg if it has a minimum effective diameter of 2.50 cm?
(b) If a force of this magnitude is applied compressively, by how much does the 25.0-cm-
long bone shorten?
Solution:
= =
2.5
2
= 1.25 = 0.0125
= = (0.125) = 4.91 × 10
1.50 × 10
N
m
=
F
A
→ F = 1.50 × 10
N
m
× 4.91 × 10 = 7.365 × 10
Thus the maximum force is 7.365 × 10
∆ = =
0.25m
1.5 × 10 N/m
× 1.50 × 10
N
m
=
0.25
100
m = 2.5mm
2. (a) Find the minimum diameter of a steel wire 18 m long that elongates no more than 9
mm when a load of 380 kg is hung on its lower end. (b) If the elastic limit for this steel
is 3 × 10 N/m , does permanent deformation occur with this load?
Solution:
=
∆
380 × 10
2
= 20 × 10 / 2 ×
9 × 10−3
18
Solve for , you will find,
= 0.000038 → = → =
0.000038
= 0.00348
≅ 3.48
Thus, ≅ 6.955
Test: =
.
= 10 / ≪ 3 × 10 N/m thus the deformation doesn’t
occur.

2. 3. Calculate the density of sea water at a depth of 1000 m, where the water pressure is
about 1.0 × 10 N/m . (The density of sea water is 1.03 × 10 kg/m at the surface, and it
bulk modulus is 2.34 × 10 N/m . )
Solution
Change in volume due to this pressure is
∆ = −
∆
= −
1.0 × 10 N/m × 1m
2.34 × 10 N/m
= −0.0043m
Thus the new volume is = 1 − 0.0043 = 0.9957 m , but the mass contained
in this volume remains unchanged, which is 1030.0
From the definition of density
= → = =
1030
0.9957
= 1.034 × 10 kg/m
ℎ ℎ =
1.034 × 10 − 1.03 × 10
1.03 × 10
× 100 = 0.4%
You may have noticed that the change is too small.
4. When water freezes, it expands by about 9%. What would be the pressure increase
inside your automobile’s engine block if the water in it froze? (The bulk modulus of ice
is 2 × 10 N/m .)
Solution:
Now, in normal conditions the water when freezes it expands by 9%, but when the water
freezes in the container, the container impose pressure on the ice preventing it from
expanding this 9%!
Thus, our initial volume is 1.09 and our final volume is after being compressed by
the container. The bulk modulus of the water is 2.00 × 10 / , thus
∆ = −
∆
= −2.00 × 10 / ×
−0.09
1.09
= 0.165 × 10 /
ℎ =
1.65 × 10 /
1.013 × 10 /
≅ 1600
This is a huge amount of pressure! The container has to be very good one to stand this
pressure.
JGood Luck J
Hassan Ashour