Torque with force applied at 90 o 800 N x 0175 m 140 Nm Physics Bell Ringer 12213 What is the torque about the shoulder if the arm is held in an abducted position at 60 degrees from the body in the frontal plane while holding a 10 kg dumbbell ID: 773490
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Torque with force applied at 90 o: 800 Nx 0.175 m 140 N•m
Physics Bell Ringer 12-2-13 What is the torque about the shoulder if the arm is held in an abducted position at 60 degrees from the body in the frontal plane while holding a 10 kg dumbbell?Assume that the mass of the arm is 6 kg, its center of mass is located 38 cm from the shoulder joint center, and the arm's total length is 80 cm.sDraw a vector diagram and show calculations. http://www.d.umn.edu/~mlevy/CLASSES/ESAT3300/PROBLEM_SETS/PDF/torque_problems.pdf
WednesdayPhysics – Final Bell Ringer, TorqueA hand exerts an upward force of 90 N on a scale at 32 cm from the joint center at the elbow. The triceps attach to the ulna at a 90° angle, and at a distance of 3 cm from the joint center. The weight of the forearm and the hand is 40 N with the forearm/hand center of gravity located 17 cm from the elbow joint center.Considering these conditions, how much force is being exerted by the triceps? (Remember that you are interested in the forces acting on the forearm system.) (answer: 733.3 N)
32 cm 17 cm90 N 40 N 3 cm X N Let CCW direction be positive and CW be negative : (40 N)(0.17 M) ‒ (90 N)(0.32 m) = ‒ (0.03 m) X X = 733 N Physics – Final Bell Ringer, Torque: Solution
Thursday, Physics Problem A charter flight company is hired to carry a load of freight using their Cessna 183 from Anchorage to Nome, AK. The aircraft specifications, along with the pilot data are given below. Where should the load of freight be placed in relaton to the “datum” in order to carry the maximum allowable weight? Assume the plane is carrying a full load of fuel. Notes: “Dry weight” means the airplane’s weight without any fuel, personnel or freight on board.The center of mass in this case serves as the axis of rotation.
datum Pilot = 200 lb. CM Freight = 746 lb. 22” 44” d=? Max allow. freight = 2950 lb. ‒ 1700 lb. ‒ 56x9 lb. ‒ 200 lb. 5 46 lb. (200)(44-22) = 546d d = 8 in. from CM From datum, d = 8+44 = 52 in.