vigorous downhill hiking, the force on the knee cartilage (the medial and latera
ID: 2256859 • Letter: V
Question
vigorous downhill hiking, the force on the knee cartilage (the medial
and lateral meniscus) can be up to eight times body weight. Depending on
the angle of descent, this force can cause a large shear force on the
cartilage and deform it. The cartilage has an area of about 11.0cm2 and a shear modulus of 12.0MPa.
(not unreasonable), and if the maximum force at impact is 8.00 times
his body weight (which, of course, includes the weight of his pack) at
an angle of 12.0 with the cartilage (see the figure (Figure 1) ),
through what angle (in degrees) will his knee cartilage be deformed?
(Recall that the bone below the cartilage pushes upward with the same
force as the downward force.)
=
During vigorous downhill hiking, the force on the knee cartilage (the medial and lateral meniscus) can be up to eight times body weight. Depending on the angle of descent, this force can cause a large shear force on the cartilage and deform it. The cartilage has an area of about 11.0cm2 and a shear modulus of 12.0MPa. If the hiker plus his pack have a combined mass of 117kg (not unreasonable), and if the maximum force at impact is 8.00 times his body weight (which, of course, includes the weight of his pack) at an angle of 12.0â with the cartilage (see the figure (Figure 1) ), through what angle (in degrees) will his knee cartilage be deformed? (Recall that the bone below the cartilage pushes upward with the same force as the downward force.)
Explanation / Answer
Shear Modulus = 12MPa = 12x10^6
Area of the cartilage = 11cm^2 = 11x10^-4 m^2
Combined mass on knee cartilage = m
= 117kg * 8 = 936 kg
F = m*g
= 936 kg * 9.8 m/s^2 = 9172.8 N
F' = F * sin(12)
= 9172.8 N * sin(12) = 1907.13 N
tan(phi) = 1907.13/(11 x10^-4 * 12x10^6) = 0.1445
Angle through which knee cartilage is deformed:
phi = arctan(0.1445)
= 8.2 degrees
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