Module 3 Practice - Solutions

Physics/BioPhysics 150 Module 3 Practice: Torque and Elasticity Solutions 1. The Achilles tendon exerts a force of magnitude F = 800 N on the heel, point P, as shown in the figure below. What is the magnitude of the torque that this tendon exerts on the ankle joint? A. 29 N ∙ m B. 9.3 N ∙ m C. 25 N ∙ m D. 14 N ∙ m E. 2.7 N ∙ m Answer: D. The torque is equal to the force applied on the heel, which is 800N, multiplied by the distance between the pivot point and the heel, which is 0.036m, and finally multiplied by the sine of the angle between the force and the straight line between the pivot and heel, which is sin (30 ° ) . Therefore, the torque is given by 𝜏 = 800(0.036) sin (30 ° ) = 800(0.036) cos (60 ° ) = 14.4𝑁.

2. Two forces produce the same torque on a lever of length a distance r from the pivot point. Which statement below is true? A. The two forces must have the same magnitude B. The two forces cannot have the same magnitude C. The component of force perpendicular to the moment arm has the same magnitude in each case D. The angle between the force and moment arm must be the same in each case E. It’s is possible to determine then angle between the force and moment arm in each case Answer: C. torque is defined as 𝜏 = ? × 𝐹 = ?𝐹 ⊥ or ? ⊥ 𝐹 . The perpendicular component of the force is what matters.

From above diagram, we can determine the signs (counterclockwise: +; clockwise: -) for each torque term using the elbow joint as the pivot point. This is a good pivot point choice because the upper arm bone also exerts a force 𝐹 𝑒 on the elbow joint and we are not given this. We therefore have two unknowns, 𝐹 and 𝐹 𝑒 . We can eliminate one based on our choice of the pivot point. We chose to  τ = −𝐹 ∗ 0.025? − ? ∗ 0.15? + ? ∗ 0.3? = 0 Solve the above equation for 𝐹 . This gives us: 𝐹 = 130𝑁 ∗ 0.3? − 27.5𝑁 ∗ 0.15? 0.025? = 1395𝑁. 4. An 12 m uniform wooden plank of mass 12 kg and rests on two supports as shown in the figure below. Suppose a crate of mass 7.5 kg is placed on the plank at a distance of 3.9 m from the left end. Find the magnitude of the force exerted by the left support on the plank. A. 190 N B. 96 N C. 130 N D. 150 N E. 180 N Answer: C. See the diagram below. Plank Plank

From above diagram. We have the torque equation using the right support as the pivot.  τ = ? ??𝑎?? ∗ (6? − 2?) + ? 𝑐?𝑎?𝑒 ∗ (3? + 7? − 3.9?) − ? ∗ 7? = 0 Note the sign of each of the torques in the equation above. Solve for n we have: ? = ? ??𝑎?? ∗ (6? − 2?) + ? 𝑐?𝑎?𝑒 ∗ (3? + 7? − 3.9?) 7? = 12?𝑔 ∗ 9.8𝑁/?𝑔 ∗ 4? + 7.5?𝑔 ∗ 9.8 𝑁/?𝑔 ∗ 6.1? 7? = 130 𝑁. 5. A surgeon is able to clamp down a needle firmly by using a suture needle holder, as shown in the figure below. If the thumb and the finger each squeeze with apply a force of magnitude |𝐹 ⃗ 𝑇 | = |𝐹 ⃗ 𝐹 | = 100𝑁 a horizontal distance 𝐷 1 = 0.63? from the pivot point P , and the needle is a horizontal distance 𝐷 2 = 0.37? away from d as shown in the figure below, where d is a positive constant, what is the magnitude of the force 𝐹 ?𝑎? that each jaw exerts on the needle? A. 170 N B. 59 N C. 340 N D. 63 N E. 27 N Answer: A

7. You come across two pumpkins balanced on a slab of wood, which is assumed to be massless and which is being held up by a support at the center of the slab. This system is static, and ? 1 ≠ ? 2 . Which of the following must be false: I. The force of gravity on each of the two pumpkins is the same as the other. II. The torques from the two pumpkins at the center of the piece of wood are equal in magnitude. III. The normal force exerted by the support is equal in magnitude to the sum of the gravitational forces on the pumpkins. IV. The sum of the forces on the wood in the vertical direction is zero. V. The pumpkins’ mass ratio is ? 1 ? 2 = ? 2 ? 1 . A. IV only B. II and III C. IV and V D. II and V E. I only Answer: E Given that the system is static we have that,  τ = τ 1 − τ 2 = ? 1 ∗ ? 1 ∗ 𝑔 − ? 2 ∗ ? 2 ∗ 𝑔 = 0 From this equation we can obtain ? 1 ? 2 = ? 2 ? 1 , ?ℎ𝑖? ????? ?? ?ℎ𝑎? ? 𝑖? ???????. In addition, since ? 1 ≠ ? 2 we conclude that ? 1 ≠ ? 2 . To get the forces that the masses exert on the slab we multiply the masses by g, then tells us that 𝐹 1 ≠ 𝐹 2 .

The net torque equation also allows to conclude that τ 1 = −τ 2 so |τ 1 | = |τ 2 | , which means statement II is correct. Since the support is carrying both of the masses then statement III is correct. This also allows us to conclude that IV is true. Therefore, only statement I is false. 8. Miss Samantha Marie Buckaroo, the guinea pig of mass 960 g, is on a mission to reach the tomatoes that were left on edge of the kitchen counter. To do this, she leans a ladder (so that it makes an angle of 𝛽 = 30 ∘ with the floor) against the counter wall and to climb up the ladder. What is the force 𝐹 ⃗ ?𝑎?? that the counter wall exerts on the ladder when Samantha is a distance ? = 1 3 ? up the ladder? The total length of the ladder is 1m. You may assume that Samantha set the ladder up such that the ladder does not slip. Assume that the wall (but not the floor) is frictionless. A. 𝐹 ?𝑎?? = 5.4𝑁 B. 𝐹 ?𝑎?? = 3.1𝑁 C. 𝐹 ?𝑎?? = 6.3𝑁 D. 𝐹 ?𝑎?? = 1.6𝑁 E. 𝐹 ?𝑎?? = 1.8𝑁 Answer: A Chose the pivot point to be where the ladder reaches the ground.

A. 1.45 m B. 0.83 m C. 0.69 m D. 1.02 m E. 0.85 m Answer: B The sum of the torques using toes as the pivot point Is given by:  τ = 𝑁 ∗ ? − ? ∗ ? = 0 Solving for ? then gives us, ? = 𝑁 ∗ ? ? = 0.665? ∗ 1.25? ? = 0.665 ∗ 1.25? = 0.831?. 10. The solid spheres shown below are made from the same, uniform material, and the mass of sphere 2 is twice that of sphere 1. The rod connecting the spheres is massless. At which location could you place a pivot such that the system remains in static equilibrium? The center of sphere 1 is located at ? = 0? and the center of sphere 2 is located at ? = 1?. Sphere 1 Sphere 2

A. It’s not possible to answer this question without knowing the mass of each sphere. B. ? = 1? C. ? = 0.5? D. ? = 0.67? E. ? = 0.33? Answer: D To determine the location of the pivot point such that the rod remains static amounts to finding the center of mass ? 𝑐? the system by using the location of small sphere to be the origin of our x coordinate. ? 𝑐? = (0?) ∗ (? 1 ) + (1?) ∗ (2 ∗ ? 1 ) (? 1 + 2 ∗ ? 1 ) = 0.67? This tells us that the pivot point should be placed at ? = 0.67? . 11. A tendon sample of length 4.00 cm and radius 2.00 mm is found to break under a minimum force of 110 N. If instead the sample had been 2.50 cm long and with a 5.00 mm radius, what would be the minimum force required to break it? A. 35.0 N B. 75.0 N C. 150 N D. 688 N E. 480 N Answer: D The stress at the breaking point is the same for both scenarios, so 𝜎 1 = 𝜎 2 , and from this 𝐹 1 𝐴 1 = 𝐹 2 𝐴 2 , where the subscript 1 corresponds to the original sample and 2 for the new one. Therefore, we have 𝐹 2 = 𝐴 2 ∗𝐹 1 𝐴 1 = π∗? 2 2 ∗𝐹 1 π∗? 1 2 = 688𝑁 12. Studies with human bones demonstrate that they behave elastically, for deformations smaller than 0.5%. Find the compressive force at the elastic limit for an adult humerus which is 0.2 m in length with a cross-sectional area of 5.0 cm 2 . The Young’s modulus for a human humerus under a compressive force is 9.0 × 10 9 𝑁 ⋅ ? −2 . Note that 1 kN = 1000 N. A. 5.17 kN

radius of 1.1 mm. What must be the radius of the steel wire if it is to stretch the same distance as the aluminum wire, so that the two wires maintain equal lengths after the masses are attached? The Young's modulus for aluminum is 0.7 × 10 11 N ∙ m −2 and for steel it is 2.0 × 10 11 N ∙ m −2 . You may assume each wire behaves elastically in this region. A. 0.65 mm B. 1.1 mm C. 2.2 mm D. 0.32 mm E. 2.5 mm Answer: A. The Young’s modulus is given by the equation 𝑌 = 𝐹/𝐴 ?/𝐿 where x is the length that the material has stretched, F is the force on the material, A is the surface area of the material, and L is the length of the material. We are given that F, x, L are all the same for the two wires. Hence 𝑌 ??𝑒𝑒? = 𝐹/(𝜋? ??𝑒𝑒? 2 ) ?/𝐿 𝑌 𝑎??? = 𝐹/(𝜋? 𝑎??? 2 ) ?/𝐿 Taking the ratio of the two, we have 𝑌 𝑎??? 𝑌 ??𝑒𝑒? = ( ? ??𝑒𝑒? ? 𝑎??? ) 2 𝑌 𝑎??? 𝑌 ??𝑒𝑒? ? 𝑎??? 2 = ? ??𝑒𝑒? 2 Plugging in the numbers to solve for ? ??𝑒𝑒? , we have 0.65mm. 15. A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 N to 0.175 N. The length and radius of th e collagen are 9.00 cm and 0.120 cm, respectively; and the Young’s modulus is 5.25 × 10 6 N∙m − 2 . If the stretching obeys Hooke’s law, what is the spring constant , ?, for the collagen? A. 128 N∙m −1 B. 264 N∙m −1 C. 196 N∙m −1

D. 300 N∙m −1 E. 517 N∙m −1 Answer: B Rearranging the equation Young’s modulus equation 𝐹 𝐴 = 𝑌 Δ𝐿 𝐿 , to 𝐹 = 𝐴𝑌 𝐿 Δ𝐿 . Gives us an equation that looks similar Hooke’s law equation, namely, 𝐹 = ?Δ𝐿 . Therefore, we can say that ? = 𝐴𝑌 𝐿 = π? 2 𝑌 𝐿 = 𝜋∗(0.12 2 ×10 −4 ? 2 )∗(5.25 × 10 6 𝑁) 9×10 −2 ? = 264 𝑁 ⋅ m −1 16. Studies with human bones demonstrate that they behave elastically, for deformations smaller than 0.5%. Find the compressive force at the elastic limit for an adult humerus which is 0.200 m in length with a cross-sectional area of 5.00 cm 2 . The Young’s modulus for compressive force of human humerus is 9.00 × 10 9 N∙m −2 . Note that 1 kN = 1000 N . A. 5.17 kN B. 14.2 kN C. 22.5 kN D. 175 kN E. 549 kN Answer: C Young’s modulus equation: 𝐹 𝐴 = 𝑌 Δ𝐿 𝐿 can be transformed to 𝐹 = 𝑌𝐴 Δ𝐿 𝐿 = 9 × 10 9 𝑁 ⋅ ? −2 ∗ 5?? 2 ∗ 0.5 × 10 −2 = 22.5 ?𝑁 Note that in above equation, Δ𝐿 𝐿 is the max deformation, which is 0.5%. 17. When the biceps muscle contracts, it exerts a force on the distal biceps tendon, causing it to stretch. Model the tendon as a cylinder of uniform material and radius ? ? that has and initial length 𝐿 (remember that the cross-sectional area of a cylinder is a circle of with area 𝜋? 2 ). If the force 𝐹 0 that the bicep exerts on the tendon causes the tendon to stretch a distance Δ𝐿 0 . How much would a tendon of radius 3? 0 2 and initial length 3𝐿 0 2 stretch in response to a force 3𝐹 0 2 ? A. 9Δ𝐿 0 /4 B. Δ𝐿 0

Answer: D. The Young’s modulus equation: 𝐹 𝐴 = 𝑌 Δ𝐿 𝐿 . Since we do not know the 𝑌 we can solve for it using the information from the problem and the graph. We can then use this to determine the Δ𝐿 for the new rod. Y = 𝐹 ∗ 𝐿 A ∗ ΔL = (40?𝑔) ∗ (9.81 𝑁 ∙ ?𝑔 −1 ) ∗ 1? π(10 −3 ?) 2 ∗ (0.002?) = 6.245 × 10 10 𝑁 ∙ ? −2 Now using this to determine the final length of the compressed new rod, Δ𝐿 = FL AY Plugging in the values given in the problem give us: Δ𝐿 = (500?𝑔)∗(9.81 𝑁∙?𝑔 −1 )∗(.01?) π(0.01?) 2 ∗(6.245×10 10 𝑁∙? −2 ) = 25 × 10 −6 ? = 25 × 10 −4 ?? So the new length is 𝐿 ?𝑒? = 𝐿 − Δ𝐿 = 10?? − 25 × 10 −4 ?? = 9.9975 ?? .