CATAGORY: Mechanics QUESTION: A 50kg high wire artist carries a weighted pole as illustrated. Her center of gravity is 80cm above the high wire and the two 23kg masses are located 120cm vertically below the wire. Find the vertical location of the center of gravity of the walker and the weighted pole relative to the high wire. Is this case of stable or unstable equilibrium.

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_ The y component of the center of mass Y is calculated by approximating the system as being made of three components. The first the 50kg wire walker whose center of mass is 80cm ABOVE the wire and two 23kg masses 120cm BELOW the wire. The y center of mass formula for a three component system is _ m(1) Y(1) + m(2) Y(2) + m(3) Y(3) (1) Y = ---------------------------------- m(1) + m(2) + m(3) Where m(1), m(2) and m(3) are the masses of the parts and Y(1), Y(2) and Y(3) are the positions of the centers of mass of each of the constitutient parts. Now consider the figure below. \||/ {..} 50kg \/ /##\ ---- 80cm | ## | + +O++++O+ + + || + + \/ + ________________________________________ 0cm__________________________ + + + + [ ] 23kg ---- -120cm 23kg [ ] Let Y(1) = 80cm , m(1) = 50kg Y(2) = -120cm , m(2) = 23kg Y(3) = -120cm , m(3) = 23kg Note the minus sign on the positions of the 23kg masses. They are negative because they are below the wire. Then _ 50kg 80cm + 23kg(-120cm) + 23kg(-120cm) 1520 kg cm (2) Y = --------------------------------------- = - ----------- 50kg + 23kg + 23kg 96 kg = -15.8cm ******************* * _ * * Y = -15.8cm * * * ******************* The system is stable because the center of gravity is below the wire.

CATAGORY: Mechanics QUESTION: A patient is in neck traction with skull calipers, as illustrated. Finf the maximum value of the suspended mass that will cause no tension or other force in the neck given that the mass of the head is 4.2 kg and the coefficient of friction between the head and the bed is o.20. (The angle between the horizontal and the force is 15 degrees.)

In this problem we balance the component of gravity force on the head that makes the head tend to slide down the pillow against the tension provided by the hanging weight and the force of friction. Lets call the positive direction DOWN the slope of the pillow. The component of gravity force in this direction is F(gravity) Pulley + + O + _ HEAD + \ + ____/ \_ __ + |- + // (X V \ + | - ||| | + | - ||| \_/ / [ ] | - \\\____ \ | - \ \ T = mg | - | - | ( - --------------------------------- 15 deg. (1) F(gravity) = M(head) g sin(15 deg.) where M(head) = 4.2kg is the mass of the head and g = 9.8m/s^2 is the acceleration due to gravity. The force of friction, F(friction) is given by the normal force, F(normal), times the coefficient of friction, k, which is equal to 0.20 in this problem. Thus (2) F(friction) = - F(normal) k or since F(normal) = M(head) g cos(15deg.) (3) F(friction) = - M(head) g cos(15deg.) k Note the minus sign means that friction operates in a direction UP the slope of the pillow. We now balance the "upward" and "downward" forces of gravity, friction and tension (which we write as -T since it works up the slope of the pillow). Thus we have (4) M(head) g sin(15 deg.) - M(head) g cos(15deg.)k - T =0 Solving for T gives (5) T = M(head) g sin(15 deg.) - M(head) g cos(15deg.)k The tension is provided by a hanging mass of weight mg so (6) T = mg or (7) m = T/g This gives us (8) m = M(head)sin(15 deg.) - M(head)cos(15deg.) k = 4.2kg sin(15 deg.) - 4.2kg cos(15 deg) (0.2) = .276 kg ~ .28kg *************** * * * m = .28kg * * * ***************

CATAGORY: Mechanics QUESTION: The diagram shows a forearm supporting a shotput whose mass is 5.44kg. Using the dimensions on the diagram find the force exerted by the biceps and the force on the elbow joint. ( Mass of forearm is 1.2 kg) ( Distance from elbow to center of gravity is 0.15m) (from center of gravity to end of hand is 0.2m)( The width of the bicep is 0.050m)

I see this problem as being like a hanging sign problem in that we must have the sum of the forces equal zero AND the sum of the torques about the elbow equal to zero. First lets look at the net torque. The torque of the muscle on the forearm is T(0.05m) where T is the tension in the muscle. | | | | | | | | |Bicep | | |Muscle | | | | \ / center of | \ / gravity Shotput | V | { } Elbow |----|--------|---------------| 0.05m | 0.10m | 0.20m | 0.05m 0.15m 0.35m Total distance from elbow The torque on the forearm due to the weight of the forearm is -mg(0.15m) and the torque on the forearm due to the shotput is -Mg(0.35m) where m is the mass of the forearm, M is the mass of the shotput and g is the acceleration due to gravity = 9.8m/s^2. A minus sign is placed on the last two torques because they cause a torque opposite to the bicep torque. This gives us (1) T(0.05m) - mg(0.15m) - Mg(0.35m) = 0 Solving for T(0.05m) and by dividing (1) by (0.05m) gives (2) T = 3mg + 7Mg Substitution of the values for m,M and g gives us. (3) T = 3(1.2 kg)9.8m/s^2 + 7(5.44kg)9.8m/s^2 = 408.5 ~ 410N ************ * * * T = 410N * * * ************ Now we look at the net force on the forearm which must sum to zero. This is made of three downward forces (i) the gravitational pull on the forearm, (ii) the weight of the shotput pressing down on the arm AND (iii) the DOWNWARD reaction force of the elbow on the forearm (-F) and one upward force T from the bicep muscle. Thus we have (4) T - mg - Mg - F = 0 Solving for F gives (5) F = T - mg - Mg = 410N - 1.2kg(9.8m/s^2) - 5.44(9.8m/s^2) = 344.9N ~ 345N ************ * * * F = 345N * * * ************