Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 12, Problem 12P
To determine
To Calculate: The force that must exert when holding a 10-lb weight out to the side at arm’s length.
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Students have asked these similar questions
A person bending forward to lift a load "with his back" (Figure a) rather than with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots
mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to
understand why back problems are common among humans, consider the model shown in Figure b, of a person bending forward to lift a W-195-N object. The spine and upper
body are represented as a uniform horizontal rod of weight W-295 N pivoted at the base of the spine. The erector spinalls muscle, attached at a point two-thirds of the way up the
spine, maintains the position of the back. The angle between the spine and this muscle is 12.0°
Back muscle
Pivot
R₂
T120
T
W
W₂
0
(a) Find the tension in the back muscle.
KN
D
(b) Find the compressional force in the spine. (Enter the magnitude.)
KN
.As a part of his daily workout routine, he lifts
10-kg dumbbells on each hand. His hands
and forearms weigh 4 kg each. If the length
of each of his forearms and hands are 0.5 m,
determine the force exerted by his muscles?
Assume that the center of gravity of the
forearms are in the middle.
Assuming a single muscle group (biceps) is responsible for carrying a ball with a weight of 5 kg as shown in the image below, what must be the force generated by the biceps so that the elbow is in static equilibrium? The weight of the forearm is 2 kg. The distance between the bicep attachment and elbow is shown as AO = 5 cm, the distance between the center of gravity of the forearm and elbow is BO = 20 cm, and the distance between the ball and elbow is CO = 40 cm.
Chapter 12 Solutions
Physics for Scientists and Engineers
Ch. 12 - Prob. 1PCh. 12 - Prob. 2PCh. 12 - Prob. 3PCh. 12 - Prob. 4PCh. 12 - Prob. 5PCh. 12 - Prob. 6PCh. 12 - Prob. 7PCh. 12 - Prob. 8PCh. 12 - Prob. 9PCh. 12 - Prob. 10P
Ch. 12 - Prob. 11PCh. 12 - Prob. 12PCh. 12 - Prob. 13PCh. 12 - Prob. 14PCh. 12 - Prob. 15PCh. 12 - Prob. 16PCh. 12 - Prob. 17PCh. 12 - Prob. 18PCh. 12 - Prob. 19PCh. 12 - Prob. 20PCh. 12 - Prob. 21PCh. 12 - Prob. 22PCh. 12 - Prob. 23PCh. 12 - Prob. 24PCh. 12 - Prob. 25PCh. 12 - Prob. 26PCh. 12 - Prob. 27PCh. 12 - Prob. 28PCh. 12 - Prob. 29PCh. 12 - Prob. 30PCh. 12 - Prob. 31PCh. 12 - Prob. 32PCh. 12 - Prob. 33PCh. 12 - Prob. 34PCh. 12 - Prob. 35PCh. 12 - Prob. 36PCh. 12 - Prob. 37PCh. 12 - Prob. 38PCh. 12 - Prob. 39PCh. 12 - Prob. 40PCh. 12 - Prob. 41PCh. 12 - Prob. 42PCh. 12 - Prob. 43PCh. 12 - Prob. 44PCh. 12 - Prob. 45PCh. 12 - Prob. 46PCh. 12 - Prob. 47PCh. 12 - Prob. 48PCh. 12 - Prob. 49PCh. 12 - Prob. 50PCh. 12 - Prob. 51PCh. 12 - Prob. 52PCh. 12 - Prob. 53PCh. 12 - Prob. 54PCh. 12 - Prob. 55PCh. 12 - Prob. 56PCh. 12 - Prob. 57PCh. 12 - Prob. 58PCh. 12 - Prob. 59PCh. 12 - Prob. 60PCh. 12 - Prob. 61PCh. 12 - Prob. 62PCh. 12 - Prob. 63PCh. 12 - Prob. 64PCh. 12 - Prob. 65PCh. 12 - Prob. 66PCh. 12 - Prob. 67PCh. 12 - Prob. 68PCh. 12 - Prob. 69PCh. 12 - Prob. 70PCh. 12 - Prob. 72PCh. 12 - Prob. 73PCh. 12 - Prob. 74PCh. 12 - Prob. 75PCh. 12 - Prob. 76P
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- As a part of his daily workout routine, he lifts 10-kg dumbbells on each hand. His hands and forearms weigh 4 kg each. If the length of each of his forearms and hands are 0.5 m, determine the force exerted by his muscles? Assume that the center of gravity of the forrearms are in the middle.arrow_forwardA person bending forward to lift a load "with his back" (Figure a) rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in Figure b, of a person bending forward to lift a Wo = 215–N object. The spine and upper body are represented as a uniform horizontal rod of weight Wb = 330 N pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12.0°. (a) Find the tension in the back muscle. (b) Find the compressional force in the spine. (Enter the magnitude.)arrow_forwardA person bending forward to lift a load "with his back" (Figure a) rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in Figure b, of a person bending forward to lift a W 170-N object. The spine and upper body are represented as a uniform horizontal rod of weight W = 355 N pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12.0°. Back muscle Pivot = a R₂ T 12.0° Rx Wb Wo (i) (a) Find the tension in the back muscle. 1.114 Your response differs from the correct answer by more than 10%. Double check your calculations. kN (b)…arrow_forward
- A person bending forward to lift a load "with his back" (Figure a) rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in Figure b, of a person bending forward to lift a Wo = 195-N object. The spine and upper body are represented as a uniform horizontal rod of weight W₁ = 305 N pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12.0⁰. Back muscle R₂ T 12.0° 1T Rx Pivot a Wb Wo ..(a).Find the tension in the back muscle. 1.114 Enter a number. differs from the correct answer by more than 10%. Double check your calculations. kN…arrow_forwardA person bending forward to lift a load "with his back" (Figure a) rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in Figure b, of a person bending forward to lift a Wo = 230–N object. The spine and upper body are represented as a uniform horizontal rod of weight Wb = 405 N pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12.0°.arrow_forwardAn individual leans forwards to pick up a box of 100 N. The weight of his upper body has a magnitude of 450 N. The back is pivoting around the base of the vertebral column. Consider the back of the individual as a rigid bar that is controlled by a muscle with an angle of 12° (See picture, d = trunk-head distance = 1 m).a) Calculate the magnitude of muscle force required to lift the box.b) Calculate the magnitude of the force at the base of the vertebral column. Hints: For (a) solve the equilibrium of moments, i.e. what force is required in the muscle to balance out the moments acting around the base of the spine.For (b), solve the equilibrium of forces acting on the spine, including the muscle force you’ve just calculated, in x and y separately. There are two extra forces not shown in the diagram: x and y contact forces acting at the base of the spine. These are whatever is needed to keep the total forces acting on the spine = 0 (so the spine isn’t accelerating off in some…arrow_forward
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