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Chapter 8: Problem 43

Find the force exerted by the biceps muscle in holding a 1-L milk carton (weight \(9.9 \mathrm{N}\) ) with the forearm parallel to the floor. Assume that the hand is \(35.0 \mathrm{cm}\) from the elbow and that the upper arm is $30.0 \mathrm{cm}$ long. The elbow is bent at a right angle and one tendon of the biceps is attached to the forearm at a position \(5.00 \mathrm{cm}\) from the elbow, while the other tendon is attached at \(30.0 \mathrm{cm}\) from the elbow. The weight of the forearm and empty hand is \(18.0 \mathrm{N}\) and the center of gravity of the forearm is at a distance of \(16.5 \mathrm{cm}\) from the elbow.

Short Answer

Expert verified
Answer: The force exerted by the biceps muscle in holding the 1-L milk carton with the forearm parallel to the floor is 119.7 N.

Step by step solution

01

Draw a diagram and identify forces

First, draw a diagram of the forearm with the elbow bent at a right angle, indicating the positions of the hand, elbow, and the attachments of the biceps tendons. Also, indicate the weight of the milk carton, the weight of the forearm and hand, and the force exerted by the biceps muscle.
02

Set the torques equal to zero

In order for the forearm to be in equilibrium, the net torque must be zero. Torque is equal to the force multiplied by the perpendicular distance to the pivot point (in this case, the elbow). Calculate the torque due to the weight of the milk carton, the weight of the forearm and hand, and the force exerted by the biceps muscle. According to the problem, the net torque should be equal to zero.
03

Calculate torque due to the milk carton

The torque due to the milk carton can be calculated as: \(\tau_{milk} = F_{milk} \times d_{1}\) where \(F_{milk} = 9.9 \mathrm{N}\) is the weight of the milk carton, and \(d_{1} = 35.0 \mathrm{cm}\) is the distance from the elbow to the hand. So, the torque due to the milk carton is: \(\tau_{milk} = (9.9 \mathrm{N})(35.0 \mathrm{cm})\)
04

Calculate torque due to the weight of the forearm and hand

The torque due to the weight of the forearm and hand can be calculated as: \(\tau_{forearm} = F_{forearm} \times d_{2}\) where \(F_{forearm} = 18.0 \mathrm{N}\) is the weight of the forearm and hand, and \(d_{2} = 16.5 \mathrm{cm}\) is the distance from the elbow to the center of gravity of the forearm. So, the torque due to the weight of the forearm and hand is: \(\tau_{forearm} = (18.0 \mathrm{N})(16.5 \mathrm{cm})\)
05

Calculate torque due to the force exerted by the biceps muscle

The torque due to the force exerted by the biceps muscle can be calculated as: \(\tau_{biceps} = F_{biceps} \times d_{3}\) where \(F_{biceps}\) is the force exerted by the biceps muscle, and \(d_{3} = 5.00 \mathrm{cm}\) is the distance from the elbow to the biceps attachment point on the forearm.
06

Set the net torque equal to zero and solve for the force exerted by the biceps muscle

Since the net torque must be zero for equilibrium, set the sum of the torques equal to zero: \(\tau_{milk} + \tau_{forearm} - \tau_{biceps} = 0\) Substitute the expressions for the torques and solve for \(F_{biceps}\): \((9.9 \mathrm{N})(35.0 \mathrm{cm}) + (18.0 \mathrm{N})(16.5 \mathrm{cm}) - F_{biceps}(5.00 \mathrm{cm}) = 0\) Solve for \(F_{biceps}\): \(F_{biceps} = \frac{(9.9 \mathrm{N})(35.0 \mathrm{cm}) + (18.0 \mathrm{N})(16.5 \mathrm{cm})}{5.00 \mathrm{cm}}\) Calculate the force exerted by the biceps muscle: \(F_{biceps} = 119.7\, \mathrm{N}\) So, the force exerted by the biceps muscle in holding the 1-L milk carton with the forearm parallel to the floor is \(119.7 \mathrm{N}\).

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Most popular questions from this chapter

A bicycle travels up an incline at constant velocity. The magnitude of the frictional force due to the road on the rear wheel is \(f=3.8 \mathrm{N} .\) The upper section of chain pulls on the sprocket wheel, which is attached to the rear wheel, with a force \(\overrightarrow{\mathbf{F}}_{\mathrm{C}} .\) The lower section of chain is slack. If the radius of the rear wheel is 6.0 times the radius of the sprocket wheel, what is the magnitude of the force \(\overrightarrow{\mathbf{F}}_{\mathrm{C}}\) with which the chain pulls?
A house painter is standing on a uniform, horizontal platform that is held in equilibrium by two cables attached to supports on the roof. The painter has a mass of \(75 \mathrm{kg}\) and the mass of the platform is \(20.0 \mathrm{kg} .\) The distance from the left end of the platform to where the painter is standing is \(d=2.0 \mathrm{m}\) and the total length of the platform is $5.0 \mathrm{m}$ (a) How large is the force exerted by the left-hand cable on the platform? (b) How large is the force exerted by the right-hand cable?
A tower outside the Houses of Parliament in London has a famous clock commonly referred to as Big Ben, the name of its 13 -ton chiming bell. The hour hand of each clock face is \(2.7 \mathrm{m}\) long and has a mass of \(60.0 \mathrm{kg}\) Assume the hour hand to be a uniform rod attached at one end. (a) What is the torque on the clock mechanism due to the weight of one of the four hour hands when the clock strikes noon? The axis of rotation is perpendicular to a clock face and through the center of the clock. (b) What is the torque due to the weight of one hour hand about the same axis when the clock tolls 9: 00 A.M.?
The mass of a flywheel is \(5.6 \times 10^{4} \mathrm{kg} .\) This particular flywheel has its mass concentrated at the rim of the wheel. If the radius of the wheel is \(2.6 \mathrm{m}\) and it is rotating at 350 rpm, what is the magnitude of its angular momentum?
A uniform diving board, of length \(5.0 \mathrm{m}\) and mass \(55 \mathrm{kg}\) is supported at two points; one support is located \(3.4 \mathrm{m}\) from the end of the board and the second is at \(4.6 \mathrm{m}\) from the end (see Fig. 8.19 ). What are the forces acting on the board due to the two supports when a diver of mass \(65 \mathrm{kg}\) stands at the end of the board over the water? Assume that these forces are vertical. (W) tutorial: plank) [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]
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