A concentric spring consists of 2 sprigs of diameter 10mm and 4mm. The...
To find the force acting on each of the two springs in the concentric spring system, we can use the principle of superposition. According to this principle, the total force acting on the composite spring is equal to the sum of the forces acting on each individual spring.
Let's denote the force acting on the larger diameter spring (10mm) as F1 and the force acting on the smaller diameter spring (4mm) as F2.
According to Hooke's Law, the force exerted by a spring is directly proportional to its displacement. Since the two springs are in series, they have the same displacement. Therefore, the force exerted by each spring can be given as:
F1 = k1 * x
F2 = k2 * x
where k1 and k2 are the spring constants of the larger and smaller diameter springs, respectively, and x is the displacement.
Given that the net force acting on the composite spring is 5000N, we have:
F1 + F2 = 5000N
Now, let's consider the relationship between the spring constants and the diameters of the springs. The spring constant is inversely proportional to the square of the diameter:
k1/k2 = (d2/d1)^2
where d1 and d2 are the diameters of the larger and smaller diameter springs, respectively.
Substituting the given values, we have:
k1/k2 = (10mm/4mm)^2 = 6.25
Now, we can solve the equations simultaneously to find the forces acting on each spring.
F1 + F2 = 5000N
F1 = 6.25F2
Substituting the second equation into the first equation, we get:
6.25F2 + F2 = 5000N
7.25F2 = 5000N
F2 = 5000N / 7.25 ≈ 689.7N
Substituting this value into the second equation, we get:
F1 = 6.25 * 689.7N ≈ 4310.3N
Therefore, the force acting on the larger diameter spring is approximately 4310.3N and the force acting on the smaller diameter spring is approximately 689.7N.
Hence, the correct answer is option 'C': 689.7N and 4310.3N.
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