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IE 665 Solution: Homework on Biomechanics 1. Three forces F1 , F2 , and F3 of magnitude 40N, 100N and 20N are acting on a point. The directions of the forces are 0, -30 and +120 degrees from the positive X axis. Determine the magnitude and direction of the resultant force. Y F3y F3=20 N 120o F1=40 N F2x X -30o F3x F2y F2=100 N F1x = F1.cos(0) = 40*1 = 40N F1y = F1.sin(0)= 40*0 = 0 N F2x = F2.cos(-30) = 100*(.866) = 86.6N F2y = F2.sin(-30)= 100*(-.5) = -50 N F3x = F3.cos(120) = 20*(-.5) = -10N F3y = F3.sin(120)= 20*(.866) = 17.32 N ΣFx =F1x +F2x +F3x = 40+86.6-10 = 116.6 N ΣFy =F1y +F2y +F3y = 0-50+17.32 = -32.68 N Resultant force F2 = (ΣFx )2 +(ΣFy )2 =116.62 +(-32.68)2 = 14663.54 F = 121.09 N Angle of the resultant force from x axis θ= tan-1 (ΣFy /ΣFx ) = tan_1 (-32.68/116.6) = -15.66o Thus the resultant of the three forces will have a magnitude of 121.09 N and it will direct at an angle 15.66o clockwise for positive X direction. 2. The applied force F1 and F2 are acting as shown below. (a)What is the magnitude of Fc? (b) What are the magnitudes of Fx and Fy at the hinge? Resolving the inclined force F1 into horizontal and vertical directions F1x = 300.cos(-30) = 300(0.866) = 259.8 N F1y = 300.sin(-30)= 300(-.5) = -150 N (negative sign indicates that this component is directed downward) For equilibrium, taking a moment about the hinge and setting it to zero, Fc*0.2 – 20(0.5) – 150*1 = 0, or Fc = 160/0.2 = 800 N Note that F1x , Fx and Fy are all passing through the hinge, so not contributing any moment at the hinge. For equilibrium, ΣFx = 0, thus F1x - Fx = 0, or Fx = F1x = 259.8N Also for equilibrium ΣFy = 0: Fy +Fc-F2-F1y = 0, or Fy = F2+F1y-Fc = 20+150-800 = -630N Fy = -630 N means it is acting opposite to the direction assumed in our diagram, and thus in reality it is acting upward. 3. Taking a moment about point of action of force FH and equating it to zero TB*1.75 – 8*14.75 = 0 , or TB = 8*14.75/1.75 = 67.43 lb To balance the forces in X direction: FH = TB-8 = 67.43-8 = 59.43 lb 4. Biceps muscle force FB is acting in a inclined way. The horizontal and vertical components of FB are FBx = FBcos(75) = 0.259FB (horizontal and towards left) FBy = FBsin(75) = 0.966FB (vertical upward) For static equilibrium, we take a moment about the point of action of the force FB and equate it to zero. i.e., FA*2 -5*12 = 0, Or, FA = 30 lb (Ans) (Note that all other forces, FC, FBx and FBy pass through the point about which the moment is taken and thus have no moment around that point) Also for static equilibrium, ΣFy = 0, FBy = FA+5 = 30+5= 35 lb Again FBy = 0.966FB, so, FB = FBy / 0.966 = 35/0.966 = 36.23 lb (Ans) And then, FBx = 0.259FB = .259* 36.23 = 9.38 lb Also, for static equilibrium, ΣFx = 0, FC= FBx = 9.38 lb (Ans). 5. We solved the low back compression forces with the following data: Load in hand mgL= 450 N Upper body weight mg BW = 350 N E = lever arm of erector spinae muscle =6.5 cm h = distance of the load from L5/S1= 30 cm b = distance of the upper body center of gravity from L5/S1 joint =20 cm Upper body angle with horizontal = 55ο The calculated results were: Spine compression force at L5/S1 disc: Fc = 3810 N Shear force L5/S1 disc: Fs = 459 N Back Muscle Force FMUSC = 3154 N Figure below shows the 3DSSPP application to for the above problem. The torso angle is kept at 55o , other joint angles are approximated to the above figure. Hand load is 450 N. The output of the program is compared with the calculated values in the table below. Spine compression (N) Spine shear (N) Back muscle force (N) 3DSSPP 4613 454 3668 Calculated value 3810 459 3154 Percent difference (%) 17% 1% 14% The maximum difference was in the spine compression force, which was 17% more than the calculated value in the class. Some difference was expected, because of the difference in anthropometry (body size and proportions), and the body angles.