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While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylinder of bone with a diameter of 2.3 cm and ignoring any shear forces. Human bone can be compressed with approximately 1.7 × 108 N/m2 before breaking. A man with a mass of 80 kg falls from a height of 3 m. Assume his acceleration once he hits the ground is constant. For these calculations, g = 10 m/s2.
Part A: What is his speed just before he hits the ground?
Part B: With how much force can the "leg" be compressed before breaking?
Part C: If he lands "stiff legged" and his shoes only compress 1 cm, what is the magnitude of the average force he experiences as he slows to a rest?
Part D: If he bends his legs as he lands, he can increase the distance over which he slows down to 50 cm. What would be the average force he experiences in this scenario?
Part E: Dyne is also a unit of force and 1 Dyn= 10−5 N. What is the maximum a bone can be compressed in Dyn/cm2?
Part F: Which of the following is the reason that we would recommend that the man bend his legs while landing from such a fall?
a. Bending his legs allows him to push back up on the ground and negate some of the effects of the force applied by the ground.
b. Bending his legs decreases the speed at which he hits the ground, thus decreasing the force applied by the ground.
c. Bending his legs decreases his overall change in momentum, thus decreasing the force applied by the ground.
d. Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Respuesta :

onyebuchinnaji

(a) The speed of man before he hits the ground is 7.75 m/s.

(b) The maximum force the leg can be compressed before breaking is 1.207 x 10⁵ N.

(c) The average force experienced by the man when his shoe compresses by 1 cm is 2.402 x 10⁶ N

(d) The average force experienced by the man when he bends his leg by 50 cm is 4,804.8 N.

(e) The maximum force in Dyn/cm² the bone can be compressed is 1.7 x 10⁹ dyn/cm².

(f) Bending his legs decreases the speed at which he hits the ground, thus decreasing the force applied by the ground.

The given parameters;

  • diameter of the cylinder, d = 2.3 cm
  • radius of the cylinder, r = 1.15 cm = 0.015 m
  • maximum stress supported by bone, τ = 1.7 x 10⁸ N/m²
  • mass of the man, m = 80 kg
  • height of the man, h = 3 m

(a)

Apply the principle of conservation of mechanical energy to determine the speed of man before he hits the ground;

[tex]K.E = P.E\\\\\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 10\times 3} \\\\v = 7.75 \ m/s[/tex]

(b)

The maximum force the leg can be compressed before breaking is calculated as follows;

F = τA

where;

  • A is the cross-sectional area of the leg

A  = πr²

A = 3.142 x (0.015)²

A = 0.00071 m²

F = ( 0.00071 m²) x (1.7 x 10⁸ N/m²)

F = 1.207 x 10⁵ N.

(c)

The average force experienced by the man when his shoe compresses by 1 cm is calculated as follows;

the acceleration of the man over the 1 cm;

[tex]v^2 = u^2 + 2a\Delta h\\\\a = \frac{v^2 -u^2}{2\Delta h} \\\\a = \frac{7.75^2 -0}{2\times 0.01} \\\\a = 30,031.3 \ m/s^2[/tex]

The average force experienced by the man;

F = ma

F = 80 x 30,031.3

F = 2.402 x 10⁶ N

(d)

The average force experienced by the man when he bends his leg by 50 cm is calculated as follows;

the acceleration over the 50 cm;

[tex]v^2 = u^2 + 2a\Delta h\\\\a = \frac{v^2 - u^2}{2\Delta h} \\\\a = \frac{7.75^2 -0 }{2\times 0.5} \\\\a = 60.06 \ m/s^2[/tex]

The average force experienced by the man;

F = 80 x 60.06

F = 4,804.8 N.

(e)

The maximum force in Dyn/cm² the bone can be compressed is calculated as;

[tex]F = 1.7\times 10^8 \frac{N}{m^2} \times \frac{1 \ dyn}{10^{-5} \ N} \times \frac{(1m)^2}{(100\ cm)^2} \\\\F = 1.7 \times 10^9 \ dyn/cm^2[/tex]

(f)

Bending his legs decreases the speed at which he hits the ground, thus decreasing the force applied by the ground.

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a) v = 7.75 m/sec

b) [tex]\rm F = 1.207\times 10^5[/tex] N

c) [tex]\rm a = 3003.125\;m/sec^2[/tex]

d) F = 4804.8 N

e) [tex]\rm F = 1.7 \times 10^9 \; dyn/cm^2[/tex]

f) The correct optoion is b) Bending his legs decreases the speed at which he hits the ground, thus decreasing the force applied by the ground.

Given :

Diameter of the cylinder, d = 2.3 cm

Radius of the cylinder, r = 1.15 cm = 0.015 m

Maximum stress supported by bone, [tex]\rm \tau = 1.7\times 10^8\;N/m^2[/tex]

Mass of the man, m = 80 Kg

Height of the man, h = 3 m

Solution :

a) Applying conservation of mechanical energy

KE = PE

[tex]\rm \dfrac{1}{2} m v^2 = mgh[/tex]

[tex]\rm v^2 = 2gh[/tex]

[tex]\rm v = \sqrt{2gh}[/tex]

[tex]\rm v = \sqrt{2\times 10 \times 3}[/tex]

v = 7.75 m/sec

b) Now we know that,

[tex]\rm F = \tau A[/tex]   ---- (1)

[tex]\rm A = \pi\times (0.015)^2 = 0.00071\;m^2[/tex]

Now put the value of A and [tex]\tau[/tex] in equation (1),

[tex]\rm F = (0.00071)(1.7\times 10^8)[/tex]

[tex]\rm F = 1.207\times 10^5[/tex] N

c) We know that,

[tex]\rm v^2=u^2+2a\Delta h[/tex] ---- (2)

Given that [tex]\rm \Delta h = 0.01\;m[/tex].

Now put the values of v, u and [tex]\rm \Delta h[/tex] in equation (2),

[tex]\rm a = \dfrac{7.75^2-0}{2\times 0.01}[/tex]

[tex]\rm a = 3003.125\;m/sec^2[/tex]

d) We know that,

[tex]\rm v^2=u^2+2a'\Delta h'[/tex]  ---- (3)

Given that [tex]\rm \Delta h' = 0.5[/tex].

Now put the values of v, u and [tex]\rm \Delta h'[/tex] in equation (3),

[tex]\rm a' = \dfrac{7.75^2-0}{2\times 0.5}[/tex]

[tex]\rm a' = 60.06\;m/sec^2[/tex]

Now,

F = ma                  (Newton's second law)

[tex]\rm F = 80\times 60.06[/tex]

F = 4804.8 N

e) Maximum force in [tex]\rm Dyn/cm^2[/tex]

[tex]\rm F = 1.7 \times 10^8 \times \dfrac{10^5}{10^4}\;dyn/cm^2[/tex]

[tex]\rm F = 1.7 \times 10^9 \; dyn/cm^2[/tex]

f) The correct optoion is b) Bending his legs decreases the speed at which he hits the ground, thus decreasing the force applied by the ground.

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