Forces and equilibrium - Orthopaedic Biomaterials in Research and Practice - page 20

PROBLEM 1.10
In Figure 1.15, the force that would tend to cause the screw and proximal
fragment (head) to telescope is

A. 1750 N

B. 1625 N

C. 615 N

D. 40 N ∙ m

E. 40 N

ANSWER:
Graphic. See Figure 1.16. The best answer is B. The device in question
is a Richards-type sliding nail-plate. It has an included angle of 145°,
its distal portion is positioned 7° from vertical, and the joint reaction
force, equal to 2.55 times standard body weight, is inclined at 12° to
the vertical (see Ref. 8 for a full discussion of hip mechanics). By con-
struction, as in Problem 1.7, the component parallel to the nail (the
telescoping force) is found to have a magnitude R T = 1650 N. Answer
C is the magnitude of the cutting out force R C , which is orthogonal to
R T . Answer D is clearly wrong, since it is a moment rather than a force
magnitude.
Numeric .

R
R

= 1750 × cos( 90 ° ° 12°

°= 683.8 N

−− = 55 ) 1610 1 . N

T

= 1750 × sin 23

C

12 º

=

R 2.5 B W
= 1750 N

145 º

1 cm = 500 N
1 cm = 4 cm

7 º

FIGUre 1.15 nail-plate fixation of intertrochanteric fracture
(Problem 1.10).