BASIC BIOMECHANICS
8th Edition
ISBN: 9781259913877
Author: Hall
Publisher: RENT MCG
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 13, Problem 5IP
Two people push on opposite sides of a swinging door. If A exerts a force of 40 N at a perpendicular distance of 20 cm from the hinge and B exerts a force of 30 N at a perpendicular distance of 25 cm from the hinge, what is the resultant torque acting at the hinge, and which way will the door swing? (Answer: Th = 0.5 N-m; in the direction that A pushes)
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Rotate the ball horizontally on an 80 cm long non-stretchable cord with angular
velocity of 3 s^(-1). After ten seconds of clockwise rotation, the cord breaks. At what
speed and in what direction does the ball fly, if it was faced north at time t = Os?
Where and after how much time does the ball land on the ground that is 1 m below
the plaine in which we rotate the string? {Solution: v= (0.37 m/s,2.37 m/s); d=1.07 m,
t=0.447 s.) }
With the shoulder flexed at 30°, the moment arm of the deltoid muscle is 2.0 cm. Solve
for the force exerted by the deltoid muscle at the glenohumeral joint give the following
assumptions:
The deltoid is the only active muscle at the glenohumeral joint
The weight of the humerus is 48 N.
The center of gravity of the humerus is located 30 cm from the shoulder center of
rotation
STATIC EQUILIBRIUM EQUATIONS
CONSIDERING ONLY THE DELTOID MUSCLE
Fo
MA =
18 Cn
COR
B=55".
0-30°
RaF 30 cm
FG = 24 N
2. a) Label the system provided below, including the reference frame, moment arms and vector forces
with the information provided.
Internal moment arm = 4cm +0.04m
External moment arm relative
to the segment weight = 25cm 0.25m
External moment arm relative
to the load weight = 45cm 40.45m
Segment weight = 50 N
Load weight = 100 N
Lower leg segment angle relative
to horizontal plane = 45°
Quadriceps tendon angle = 45°
Axis of
rotation
MF
SW
LW
2b) Using the figure in 2a., calculate the external torque of the system relative to the normal
component of segment and load weights listed above.
2c) Calculate the amount of both the tangential component of the muscle force and the muscle force
itself required to keep this system in a state of static equilibrium.
Chapter 13 Solutions
BASIC BIOMECHANICS
Ch. 13 - Why does a force directed through an axis of...Ch. 13 - Why does the orientation of a force acting on a...Ch. 13 - A 23-kg boy sits 1.5 m from the axis of rotation...Ch. 13 - Prob. 4IPCh. 13 - Two people push on opposite sides of a swinging...Ch. 13 - Prob. 6IPCh. 13 - Prob. 7IPCh. 13 - Prob. 8IPCh. 13 - A 10-kg block sits motionless on a table in spite...Ch. 13 - Prob. 10IP
Ch. 13 - A 35-N hand and forearm are held at a 45 angle to...Ch. 13 - A hand exerts a force of 90 N on a scale at 32 cm...Ch. 13 - A patient rehabilitating a knee injury performs...Ch. 13 - A worker leans over and picks up a 90-N box at a...Ch. 13 - A man carries a 3 m, 32-N board over his shoulder....Ch. 13 - A therapist applies a lateral force of 80 N to the...Ch. 13 - Tendon forces Ta and Tb are exerted on the...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, bioengineering and related others by exploring similar questions and additional content below.Similar questions
- A) Describe the difference between strength and power during a squat. Provide a quantitative example that illustrates the difference. B) During a squat, the weight you have on your shoulders has a certain amount of gravitational potential energy. As you squat down and come to a stop at the bottom of the squat, some of that energy is lost. Where does it go?arrow_forwardThe adjacent diagram displays the joint reaction force between the acetabulum and femoral head for the right leg during single leg stance. Also shown are the abductor muscle force (AMF), the weight of the body above the level of the stance hip (W) and their corresponding moment arms with respect to the joint centre (D and D1). If the right leg weighs 1/6 of total body weight, the ratio of D1 to D is 2.4, and AMF is angled at 30 degrees relative to the vertical, calculate the: Abductor muscle force (2 marks). Hip joint reaction force (2 marks). Express both answers as multiples of total body weightarrow_forwardGiven the following ankle and knee coordinates calculate the partial internal knee angle to 1 decimal place Ankle z 0.129 m Ankle x 0.37 m Knee z 0.409 m Knee x 0.522 m Answer: re to search 8: T G B N W 8 M 9 O Pa P 4 3°C℃ Partly cloudyarrow_forward< The three main forces that act on the patella are shown on the diagram of the knee joint below. These forces are the quadriceps muscle force (FQ), the patella ligament force (FPL), and the patellofemoral joint reaction force (FPF). The angles a and ẞ are with respect to a line that is perpendicular to FPF. Assuming a = 15°, ẞ = 20°, and FQ = 3725 N, use equations for static equilibrium to calculate (a) FPL, and (b) FPF. (Hint: To solve this problem consider using a coordinate system that is aligned with the principal axes of the patella instead of the usual vertical and horizontal axes). (c) Many people assume FQ and FPL are always equal in magnitude but this is only true under certain circumstances. Under what conditions are FQ and FPL equal in magnitude? Fo FPF FP B GriffithUNIVERSITY Queensland Australiaarrow_forwardDefine the following parameters that can be assessed via isokinetic dynamometry (attempt to use your own words after reviewing data sheet): Peak Torque Time to peak torque Angle of peak torque Torque at 0.2 seconds Peak torque/body weight Total work Work fatigue (Fatigue Index) Average powerarrow_forward3 E there to search D Calculate the ankle torque during initial gait contact from the following data set to 1 decimal place. NOTE if your answer is negative put the negative sign in the answer! FZ 765 N 4 с Fx-290 N CoP z 0 m Cop x 0.51 m Ankle z 0.074 m Ankle x 0.62 m Answer Check R F 96 5 Bi T G B 6 y H N U J 8 M 1 hp к Ji C 9 9 P O alt gr O P ? ctri 910- C > pri se 1 GBP/AUD-0.41% pause # ← tock Next page 7 11:26 02/12/2023 1 end ☐ po up 5 Carrow_forwardIf the muscle fiber is stretched to 150% of muscle length and thentechnically stimulated, what would be the total force measured?arrow_forwardNonearrow_forwardney: Load (L) = 5 kgs = Effort E = Fulcrum Weight of forearm = 1.8 kgs L = Load Biceps brachii muscle Distance of load from elbow joint = 35 cm %3D Effort (E) = contraction of biceps brachii Distance of center of mass of forearm from elbow = 17 cm Distance of tendon from elbow = 4 cm A) Draw the free-body diagram to represent the forces and moments Load (L) = weight of object plus forearm B) Write the torque equation for static equilibrium Fulcrum (F) = elbow jointarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
Recommended textbooks for you
Human Physiology: From Cells to Systems (MindTap ...BiologyISBN:9781285866932Author:Lauralee SherwoodPublisher:Cengage Learning
Human Physiology: From Cells to Systems (MindTap ...
Biology
ISBN:9781285866932
Author:Lauralee Sherwood
Publisher:Cengage Learning
Chapter 7 - Human Movement Science; Author: Dr. Jeff Williams;https://www.youtube.com/watch?v=LlqElkn4PA4;License: Standard youtube license