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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 large clock has a second hand with a mass of \(0.10 \mathrm{kg}\) concentrated at the tip of the pointer. (a) If the length of the second hand is $30.0 \mathrm{cm},$ what is its angular momentum? (b) The same clock has an hour hand with a mass of \(0.20 \mathrm{kg}\) concentrated at the tip of the pointer. If the hour hand has a length of \(20.0 \mathrm{cm},\) what is its angular momentum?
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