Respuesta :
Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament, [tex]P=100\times 10^6\ N/m^2=10^8\ Pa[/tex]
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,
[tex]F=P\times A\\\\F=10^8\times \pi (2.4\times 10^{-3})^2\\\\F=1809.55\ N[/tex]
So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
To determine the maximum force that could be applied to anterior cruciate ligament (ACL) is [tex]13.3 \times 10^9 \;Newton[/tex]
Given the following data:
- Tensile strength = [tex]100 \times 10^6 \;N/m^2[/tex]
- Diameter = 4.8 mm to m = 0.0048 meter.
Radius = [tex]\frac{Diameter}{2} = \frac{0.0048}{2} = 0.0024\;meter[/tex]
To determine the maximum force that could be applied to anterior cruciate ligament (ACL):
First of all, we would find the area of the anterior cruciate ligament (ACL) by using this formula:
[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.0024[/tex]
Area = 0.007541 square meter.
Now, we can determine the maximum force:
Mathematically, force is given by the formula:
[tex]Force = \frac{Tensile \;strength}{Area} \\\\Force = \frac{100 \times 10^6 }{0.007541} \\\\Force = 13.3 \times 10^9 \;Newton[/tex]
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