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We analyzed the biceps muscle example with the angle between forearm and upper arm set at 90º . Using the same numbers as in [link] , find the force exerted by the biceps muscle when the angle is 120º and the forearm is in a downward position.

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Even when the head is held erect, as in [link] , its center of mass is not directly over the principal point of support (the atlanto-occipital joint). The muscles at the back of the neck should therefore exert a force to keep the head erect. That is why your head falls forward when you fall asleep in the class. (a) Calculate the force exerted by these muscles using the information in the figure. (b) What is the force exerted by the pivot on the head?

An erect head is shown. The weight of the head is fifty newtons. The center of gravity of the head lies in front of its support. The perpendicular distance between the support and the weight of the head is two point five centimeters. Between these forces, there is a point where a vertical force vector is shown. This force is marked as F sub J. At the back of the head, five point zero centimeters behind the support point, is a downward vector labeled F sub m.
The center of mass of the head lies in front of its major point of support, requiring muscle action to hold the head erect. A simplified lever system is shown.

(a) 25 N downward

(b) 75 N upward

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A 75-kg man stands on his toes by exerting an upward force through the Achilles tendon, as in [link] . (a) What is the force in the Achilles tendon if he stands on one foot? (b) Calculate the force at the pivot of the simplified lever system shown—that force is representative of forces in the ankle joint.

A foot of a person is shown. The ankle is slightly above the ground. There is a force in F-A on the back part of ankle, which is in upward direction. The weight of the leg is downward. The normal reaction is acting at the front foot in upward direction. The perpendicular distance between the normal reaction and the force F-A is sixteen centimeters. There is a point between these two forces where a force F-P is shown, which acts as fulcrum of the simplified lever system.
The muscles in the back of the leg pull the Achilles tendon when one stands on one’s toes. A simplified lever system is shown.

(a) F A = 2 . 21 × 10 3 N size 12{F rSub { size 8{A} } =2 "." "21" times "10" rSup { size 8{3} } `N} {} upward

(b) F B = 2.94 × 10 3 N size 12{F rSub { size 8{B} } =2 "." "94" times "10" rSup { size 8{3} } `N} {} downward

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A father lifts his child as shown in [link] . What force should the upper leg muscle exert to lift the child at a constant speed?

A leg of a person is shown. On the foot, a child is sitting. The weight of the child is ten kilograms acting downward. The center of gravity of the leg is shown at the middle part of the lower leg. The knee is acting as the pivot. The mass of the leg is marked as four kilograms. The distance of the head of the child is thirty eight centimeters from the pivot and the perpendicular distance between the center of gravity of the leg and pivot is twenty centimeters.
A child being lifted by a father’s lower leg.
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Unlike most of the other muscles in our bodies, the masseter muscle in the jaw, as illustrated in [link] , is attached relatively far from the joint, enabling large forces to be exerted by the back teeth. (a) Using the information in the figure, calculate the force exerted by the lower teeth on the bullet. (b) Calculate the force on the joint.

The masseter muscles of a jaw of a man are shown. The force F sub M is equal to two hundred newtons and is acting on the muscle in upward direction and the force F sub J is acting to the left end of the muscle downward. The span of the muscle at upper part is five centimeters. At the joint of jaw, the reaction force is downward.
A person clenching a bullet between his teeth.

(a) F teeth on bullet = 1.2 × 10 2 N size 12{F rSub { size 8{r} } ="116"`N} {} upward

(b) F J = 84 N size 12{F rSub { size 8{J} } ="84"`N} {} downward

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Integrated Concepts

Suppose we replace the 4.0-kg book in [link] of the biceps muscle with an elastic exercise rope that obeys Hooke’s Law. Assume its force constant k = 600 N/m size 12{k="600"`"N/m"} {} . (a) How much is the rope stretched (past equilibrium) to provide the same force F B size 12{F rSub { size 8{B} } } {} as in this example? Assume the rope is held in the hand at the same location as the book. (b) What force is on the biceps muscle if the exercise rope is pulled straight up so that the forearm makes an angle of 25º size 12{"25"°} {} with the horizontal? Assume the biceps muscle is still perpendicular to the forearm.

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(a) What force should the woman in [link] exert on the floor with each hand to do a push-up? Assume that she moves up at a constant speed. (b) The triceps muscle at the back of her upper arm has an effective lever arm of 1.75 cm, and she exerts force on the floor at a horizontal distance of 20.0 cm from the elbow joint. Calculate the magnitude of the force in each triceps muscle, and compare it to her weight. (c) How much work does she do if her center of mass rises 0.240 m? (d) What is her useful power output if she does 25 pushups in one minute?

A woman doing pushups is shown. The weight W of her body is acting at the middle point of the length of her body. Her palms are on the ground. The distance between the palm and the feet is one point five meters. The distance between the center of gravity and the feet is zero point nine meters. The normal reaction on her hands is acting upward.
A woman doing pushups.

(a) 147 N downward

(b) 1680 N, 3.4 times her weight

(c) 118 J

(d) 49.0 W

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You have just planted a sturdy 2-m-tall palm tree in your front lawn for your mother’s birthday. Your brother kicks a 500 g ball, which hits the top of the tree at a speed of 5 m/s and stays in contact with it for 10 ms. The ball falls to the ground near the base of the tree and the recoil of the tree is minimal. (a) What is the force on the tree? (b) The length of the sturdy section of the root is only 20 cm. Furthermore, the soil around the roots is loose and we can assume that an effective force is applied at the tip of the 20 cm length. What is the effective force exerted by the end of the tip of the root to keep the tree from toppling? Assume the tree will be uprooted rather than bend. (c) What could you have done to ensure that the tree does not uproot easily?

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Unreasonable Results

Suppose two children are using a uniform seesaw that is 3.00 m long and has its center of mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from the pivot. (a) Calculate where the second 18.0 kg child must sit to balance the seesaw. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?

a) x - 2 = 2.33 m size 12{ { bar {x}} rSub { size 8{2} } =2 "." "33"`m} {}

b) The seesaw is 3.0 m long, and hence, there is only 1.50 m of board on the other side of the pivot. The second child is off the board.

c) The position of the first child must be shortened, i.e. brought closer to the pivot.

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Construct Your Own Problem

Consider a method for measuring the mass of a person’s arm in anatomical studies. The subject lies on her back, extends her relaxed arm to the side and two scales are placed below the arm. One is placed under the elbow and the other under the back of her hand. Construct a problem in which you calculate the mass of the arm and find its center of mass based on the scale readings and the distances of the scales from the shoulder joint. You must include a free body diagram of the arm to direct the analysis. Consider changing the position of the scale under the hand to provide more information, if needed. You may wish to consult references to obtain reasonable mass values.

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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