This document is the first midterm for a general physics course for medical students. It contains 4 questions regarding vectors, torques, forces on the forearm, and stresses on bone. The solutions provide the following key details: 1) For two vectors A and B in the xz-plane, it calculates their scalar product, vector product, and the magnitude and direction of the resultant vector 3A - 2B. 2) For a diagram of a wheel, it finds the net torque on the wheel about the axle. 3) For a free body diagram of the forearm, it determines the tension in a muscle and the force at the elbow joint given applied forces. 4)

1. Faculty of Science
Phys 1301: General Physics for Medics
First Midterm
Allowed time: 90 minutes
Instructors: Dr. Hassan Ashour, Dr. Naji Al-Dahoudi
Answer the following Questions
Q1: Two vectors 𝐴 and 𝐵 are confined in 𝑥𝑧 −plane as shown in figure, find
1. The scalar product 𝐴 ∙ 𝐵
2. The vector product 𝐴 × 𝐵
3. The magnitude and direction of the resultant vector 3𝐴 − 2𝐵
Solution
The vector 𝑨 is 𝐴 = −5√2 cos 45 𝑖— 5√2 sin 45𝑘 → 𝐴 = −5𝑖 − 5𝑘
The vector B is 𝐵 = 7𝑖
1. The dot product of 𝐴 ∙ 𝐵 = −5(7) = −35 𝑢𝑛𝑖𝑡
2. The cross product of 𝐴 × 𝐵 = |
𝑖 𝑗 𝑘
−5 0 −5
7 0 0
| = −𝑗(−(−5)(7)) = −35 𝑗
3. 𝐶 = 3𝐴 − 2𝐵 = 3(−5𝑖 − 5𝑘) − 2(7𝑖) = −29𝑖 − 15𝑘
𝐶 = √292 + 152 = √1066 = 32.65 𝑢𝑛𝑖𝑡
𝜃 = 𝑡𝑎𝑛−1
(
−15
−29
) = 27.35°

2. Q2: Find the net torque on the wheel in the figure below about the axle through O if 𝑎 is
10.0 cm and 𝑏 is 25.0 cm.
Solution
We chose to the following direction
𝑐. 𝑤. 𝑔𝑒𝑡𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑠𝑖𝑔𝑛
𝑐. 𝑐. 𝑤. 𝑔𝑒𝑡𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑠𝑖𝑔𝑛
𝜏 = −10 × 0.25 𝑐. 𝑤. −9 × 0.25 𝑐. 𝑤. +12 × 0.1𝑐. 𝑐. 𝑤
𝜏 = −3.55 𝑁𝑚 𝑐. 𝑤.
Q3: A realistic free body diagram model for the forearm. In the figure below, 𝑤1 = 12 𝑁
find the tension T exerted by the muscle and the force exerted by the elbow joint E.
Solution
2
121.85
nd law
N
first law
´
=
´
‫ق‬
-12×0.35-12×0.15+Tcos10×0.05=0
solve for T
0.05 Tcos 10=12 0.5,
6
T =
0.05 cos10
Tcos 10-12-12-E =0 Tcos 10= 24+E
E = Tcos10-24=96 N

3. Q4: Assume that Young’s modulus for bone is 1.5 × 1010
N/m2
and that a bone will
fracture if more than 1.50 × 108
N/m2
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:
𝑅𝑎𝑑𝑖𝑢𝑠 = 𝑟 =
2.5
2
= 1.25𝑐𝑚 = 0.0125𝑚
𝐴𝑟𝑒𝑎 = 𝜋𝑟2
= 𝜋(0.125)2
= 4.91 × 10−4
𝑚2
1.50 × 108
N
m2
=
F
A
→ F = 1.50 × 108
N
m2
× 4.91 × 10−4
𝑚2
= 7.365 × 104
𝑁
Thus the maximum force is 7.365 × 104
𝑁
∆𝑙 =
𝑙
𝑌
𝐹
𝐴
=
0.25m
1.5 × 1010 N/m2
× 1.50 × 108
N
m2
=
0.25
100
m = 2.5mm
Good Luck 
Ashour-Dahoudi