# T elbow = - (F * R) =- ( 100 N * 0.3 meters) = - 30 Nm

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1 Biomechanics IPHY 4540 Problem Set #10 For all quantitative problems, please list all known variables, the variables that you need to find, and put a box around your final answers. Answers must include units. Show the work you did to reach the answer. 1. Work is measured in Joules which have units of N m. Torque is measured in N m. Explain how work and torque are different even though the units seem to be the same. Work can be calculated as the product of force and distance, F d. The force and distance travelled must be along the same line of action. Torque can be calculated as the product: r x F. For torque calculations, the distance r is the perpendicular distance from the line of action of the force and the center of rotation. Torque tends to cause rotation of an object. 2. A person is holding a weight (100 N) in their hand. The weight is a distance of 0.3 meters from the center of rotation of the elbow. A. If the forearm is parallel to the ground, what is the torque about the elbow due to the weight? T elbow = - (F * R) =- ( 100 N * 0.3 meters) = - 30 Nm B. If the forearm has an angle of 30 above the horizontal (hand is higher than elbow), what is the torque about the elbow due to the weight? T elbow = - (F * R) R = 0.3cos 30 = 0.26 meters T elbow = - (100 N * 0.26 m) = - 26 Nm C. If an elbow flexor muscle has a moment arm of m about the elbow, what force must it generate to hold the forearm at an angle of 50 above the horizontal with the weight in the hand (ignoring the weight of the forearm)? ΣM elbow = 0: - F w R w + F m R m = 0 where F m & R m are the muscle force and its moment arm respectively, and F w & R w are the 100N weight and its moment arm respectively. R w = 0.3 * cos 50 = 0.19 meters * F m * 0.03 = 0 F m = 633 N 5. A person is holding a weight (100 N) in their hand. The weight is a distance of 0.3 meters from the center of rotation of the elbow. A. If the forearm is parallel to the ground, what is the torque about the elbow due to the weight?

2 Torque = - (F*R) =- ( 100N * 0.3m )= - 30 Nm B. If that torque is counteracted by a single biceps muscle (call it, B1) acting with a moment arm of 2.5 cm, what is the muscle force in that biceps muscle? ΣM = 0 = - (100N * 0.3m) + (F B1 * 0.025m) F B1 = 1200N C. If the triceps muscle (an elbow extensor, antagonist to the biceps) is co- contracting and exerting a force of 1000N, how would that quantitatively affect the biceps force needed to keep the elbow stationary? Moment arm of triceps is 0.04m. We are looking for an explanation and a calculation here. The moments must sum to zero. If there is an additional extensor torque, then there must be an even greater biceps flexor torque to counteract it plus the extensor torque of the weight held in the hand. ΣM = 0 = - (100N * 0.3m) + (F B1 * 0.025m) + (1000 * 0.04) F B1 = 2800N D. If the torque is counteracted by the actions of two biceps muscles (but no triceps muscle force), explain why you can not solve for the muscle force in each biceps muscle with 100% confidence. There is one equation, ΣM = 0 but two unknowns (F B1 & F B2 ). Can t solve that with algebra. E. Again, assume no triceps force for this part. The cross sectional area of muscle B1 is twice that of muscle B2. The moment arm of muscle B2 is 1.5cm. Assuming that muscle force is proportional to cross sectional area, i.e. that muscle stress is minimized, calculate the muscle forces in B1 and B2. ΣM = 0 = - (100N * 0.3m) + (F B1 * 0.025m) + (F B2 * 0.015m) Given F B2 = 0.5 * F B1 ΣM = 0 = - (100N * 0.3m) + (F B1 * 0.025m) + (0.5* F B1 * 0.015m) ΣM = 0 = - 30N (F B1 ) F B1 = 923N, and F B2 = half that or 461N. F. List the situations (B,C,E) in order of descending joint reaction forces (JRF). Start with the greatest JRF. C, E, B

3 C has the triceps co- contraction, so larger biceps and triceps muscle forces must be opposed by JRF. B is the least, because the B1 muscle force is less than the sum of B1 +B2 in situation C. 6. A person is wearing a weight boot (150N) and doing exercises to strengthen the knee extensor muscles. The center of mass of the weight boot is 0.35 meters from the center of rotation of the knee. The person's foot + shank has a weight of 40N and its center of mass is 0.2 meters from the center of rotation of the knee. These distances are along the length of the shank. A. When the shank is perpendicular to the ground, what is the total torque about the knee? (Be sure to use a free- body diagram.) All of the moment arms are zero because the force vectors go directly through the center of rotation of the knee. As a result, the total torque is 0. B. When the shank is held at a position where it is 20 below the horizontal (foot is lower than knee), what is the torque about the knee? (Be sure to use a free- body diagram. Note that the moment arm of the weight boot about the knee changes with knee angle.) T = - (T due to shank weight + T due to weight boot) = - [(40)(0.2 cos 20)] - [(150)(0.35 cos 20)] = N*m C. If the quadriceps muscle group has a moment arm of m, what is the muscle force needed to hold the shank at 20 below the horizontal? ΣM knee = 0 = - [(40)(0.2 cos 20)] - [(150)(0.35 cos 20)] * F m,quad F m,quad = 2276 N 7. A squat is a weight lifting exercise that begins while standing upright with a barbell supported on the shoulders The knees are flexed and then extended, trying to keep the back fairly straight upright. In terms of torques, moment arms and muscle forces, why is it so much more difficult to do a full, deep squat compared to a half squat in which you only squat down part way before rising? Going down to a full squat puts the knees at the extreme of their range of motion. The knee extensor muscles cannot exert very high moments there because the moment arms are smaller and the muscles are operating at long lengths. 8. Torques act in the sagittal and frontal planes and they also act in twisting. Skiers suffer many more knee injuries than snowboarders. Explain why with special emphasis on torques.

4 Skiers have longer boards that can act as long levers and thus impose large sudden external torsional twisting torques on the leg, for example if the skier catches a ski tip on a tree or other obstacle. 9. Based on bite marks in the bones of an unfortunate triceratops, T. rex could exert a biting force of 100,000.0 N measured at the tooth. Draw a FBD where the system is the lower jaw (mandible) of the T- rex. A. If that force was exerted at a distance of 1.0 m from the jaw center of rotation, and the T. rex masseter muscle has a moment arm of 40.0cm, how much muscle force was needed? (Neglect the weight of the jaw) Sum of the M Jt = 0 = F m *R m - F b *R b ;F m = (100000N*1.0m)/0.4m = 250,000N B. What is the joint reaction force? (Again, neglect the weight of the lower jaw). Sum of the forces in the vertical = F m + F j - F b = 0; F j = - 250, ,000 = - 150,000N C. The mandible of a T- rex may weigh more than hundred pounds, is it reasonable to neglect it? 100 lbs. seems big to us but relative to the muscle forces involved it is pretty small. 100lbs. ~ 45 kg = 450 N compared to 250,000N. 10. Estimate the force in the ankle extensor tendon (i.e. Achilles) at mid- stance for a 70.0 kg person who is walking. Base your answer on data from the textbook, your notes and from your own leg. List any of these values that you estimate/assume. You need to estimate the ground reaction force, the point of force application (center of pressure) and the moment arm of the tendon. Assume that the ankle is quasi- static at midstance (sum of moments = 0). The point of this problem was to get you thinking about what forces and moment arms are involved in a real analysis of the moments at the ankle during locomotion.

5 Full credit requires accounting for the relevant forces at appropriate levels of magnitude. It was not necessary to get exactly the quantitative answer listed below. Whatever values you estimated should be substituted. weight foot = * body weight = 6.2N F gy ~ 500N, F gx ~ 0N at midstance estimated moment arms, 15 cm from point of grf vector to ankle joint, 8cm from com of foot to ankle, moment arm of Achilles tendon 5 cm. Sum of the Moments at the ankle = 0 = (F m *R m ) + (F mg foot *R foot ) (F gy *R gy ) F m = [- (6.2N*0.08)+(500N*0.15m)]/0.05m F m = 1490 N 11. Consider the following cross- country skier. In anticipation, he considers the torques exerted about his elbow when he is applying a force to his ski pole during the propulsive phase of the ski stride. His upper arm is vertical. His elbow is at an angle of 120 degrees. The pole is held at an angle of 70 degrees to the forearm. The triceps muscle has a moment arm of 4.0 cm. The force exerted along the shaft of the pole is 200Newtons. Ignore the weight of the forearm & hand. Your goal is to calculate the triceps force and the elbow joint reaction force magnitude and direction. Some information that may or may not be useful: the pole is 180cm long, the distance from the pole grip to the elbow joint is 40 cm. Assume that the triceps muscle pulls parallel to the upper arm.

6 A. But first, draw an appropriate free body diagram to help you solve parts B, C & D. You should isolate the forearm as the system. Force acting on it are: the triceps force, the vertical and horizontal pole forces and the joint reaction force. B. Solve for the triceps force. The 120 angle can be broken down into a 90 + a 30 degree angle. The 70 can be broken down into a 30 & a 40 degree angle. This allows for easier calculation of the R s & the x & y components of the pole force. ΣM = 0 = - (triceps force * 0.04) + (Fy pole * Ry) + ( Fxpole *Rx) = F T + ((200 sin 40deg) * (0.4 * cos 30 deg)) + ((200cos 40deg)* (0.4 sin 30deg)) = F T + ((200 * 0.643) * (0.4 * 0.866)) + ((200 * 0.766)* (0.4 * 0.5)) = F T + (128.6 * 0.346) + (153.2 * 0.2) = F T + (44.5) + (30.6) F T = 1879N C. Solve for the magnitude and direction of the resultant elbow joint reaction force. ΣFy = 0 = F T + (200 sin 40 deg) + F jrfy ΣFy = 0 = (200 * 0.642) + F jrfy F jrfy = - ( ) = N (- ve means down in our default coordinate system) ΣFx = 0 = (200 cos 40 deg) + F jrfy F jrfy = - (200 * 0.766) = N (- ve mean back in our default coordinate system) Pythagorean theorem, F result = sqrt ( ) = 2013N down and 4.3 degrees behind vertical D. Would the calculated muscle force be greater or smaller if you included the weight of the forearm? Smaller, the weight would exert an extensor torque, so less triceps torque would be needed. 12. It is easy for most of us to hold a glass, but a person recovering from a cervical spinal cord injury it can be very difficult and they must re- learn how to apply the needed forces. A person is standing in a bar, holding a glass with straight sides. The mass of the glass plus water is 500 grams. Coefficient of friction is 0.8. Diagram the forces that the person applies to the glass. Calculate the frictional force that the thumb must exert on the glass. Calculate the minimum grasping force that the thumb must exert on the glass. Grasping force = the force perpendicular to the surface of the glass.

7 13. You are viewing the right side of a cyclist moving from your left to right.! A. When the right crank is at the 3 o clock position, the rider applies a vertical force to the pedal of 400N. The crank is 170 mm long. What is the torque applied to the crank axle? Torque = - (F*R) = - (400N*0.17m) = - 68Nm B. At the top of the pedal cycle, i.e. 12 o clock, the rider must apply a 400N horizontal force in order to generate the torque calculated in part a. Given the drawing above, (and ignoring the weight of the foot and calf) calculate the knee torque that must be generated. Is it positive or negative and why? M knee = (400N*.45m) = +180Nm The torque (moment) is an extensor torque, because the knee must be extended in order to exert the torque on the pedal. C. Calculate the quadriceps muscle force needed. The moment arm for the quadriceps is 4.0 cm. M knee = F m *R m = 180Nm; 180Nm/0.04m =4500N

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