Basic Biomechanics
7th Edition
ISBN: 9780073522760
Author: Susan J Hall
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 1, Problem 10IP
A 0.5 kg ball is kicked with a force of 40 N. What is the resulting acceleration of the ball?
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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.
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.) }
Based on the acceleration in the above ball rolling down an inclined plane (with vo = 0 meters per second), how far would it have traveled along the inclined plane in the first six seconds of rolling?
9 meters
16 meters
25 meters
36 meters
49 meters
Chapter 1 Solutions
Basic Biomechanics
Ch. 1 - Write a brief discussion about how knowledge of...Ch. 1 - Prob. 4IPCh. 1 - Write a summary list of the problem-solving steps...Ch. 1 - Prob. 6IPCh. 1 - Step by step, show how to arrive at a solution to...Ch. 1 - Solve for x in each of the equations below. Refer...Ch. 1 - Two schoolchildren race across a playground for a...Ch. 1 - A 0.5 kg ball is kicked with a force of 40 N. What...Ch. 1 - When attempting to balance your checkbook, you...Ch. 1 - Prob. 3AP
Ch. 1 - Wendell invests 10,000 in a stock portfolio made...Ch. 1 - The hypotenuse of right triangle ABC (shown here)...Ch. 1 - In triangle DEF, side E is 4 cm long and side F is...Ch. 1 - An orienteer runs 300 m north and then 400 m to...Ch. 1 - Prob. 8APCh. 1 - Prob. 9APCh. 1 - A sailboat heads north at 3 m/s for 1 hour and...
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- A scientist was investigating if differences in the frictional work performed on a model car can change depending on its mass (in grams) and whether the car moves up or down an inclined plane. They decided to measure the amount of frictional force experienced by the model car and the distance it traveled in meters. The scientists were able to evaluate the frictional work using the following data. Mass (g) Distance (m) Force Work Done by Friction (J) car going up the incline 100 39 0.063 2.457 car going down the incline 70 39 0.2309 ? It is known that the relationship between force and distance determines the work done by friction (W+). W₁ = fd Wf work done by friction f = force d = distance Question: How much work done by friction was exerted on the car as it moved down the inclined plane? You may use a calculator. 1 2.457 9.005 11.46 16.16 PREVIOUS FINISHarrow_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_forwardWhen Galileo Galilei rolled a ball down an inclined plane, it traveled 1.0 meters in the first second, and a total of 4.0 meters in the first two seconds. What was its acceleration on this inclined plane? 2.0 meters per second2 3.0 meters per second2 4.0 meters per second2 5.0 meters per second2 6.0 meters per second2arrow_forward
- Total compressive knee forces range between O 1-2 %BW O 2-4 %BW 3-5 %BW 0.5-1.5 %BWarrow_forwardWhat is the relationship between Net force, mass and acceleration. Newton’s second law of motion?arrow_forwardWith 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 Narrow_forward
- A spring has a constant of 875 N/m. What hanging-mass will cause this spring to stretch 4.5 m?arrow_forwardWhat was the ball’s acceleration on the inclined plane described above? 2.0 meters per second2 B. 3.0 meters per second2 4.0 meters per second2 6.0 meters per second2 8.0 meters per second2arrow_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_forward
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