Download biomechanics hw sol

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.