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Answer :
Final answer:
The displacement of the object from its equilibrium position, supported by two springs at an angle, is calculated by using Hooke's Law and trigonometry, resulting in a displacement of approximately 0.1934 meters.
Explanation:
To find the displacement of the object from its equilibrium position, we need to consider both the weight of the object (w) and the spring constant (k). Since there are two identical springs at an angle θ, each spring only needs to support half the weight of the object. We can use trigonometry to find the vertical component of the restoring force from a spring, which must equal half the object's weight for the system to be in equilibrium.
The restoring force offered by a spring is given by Hooke's Law, which is F = k * x, where F is the force, k is the spring constant, and x is the displacement. The vertical component of this force is F * cos(θ). Since each spring supports half of the object's weight, we can write:
0.5 * w = k * x * cos(θ)
Substituting the given values:
0.5 * 14.4 N = 39.4 N/m * x * cos(19.1°)
x = (0.5 * 14.4 N) / (39.4 N/m * cos(19.1°))
By rearranging the above formula and solving for x, we obtain the displacement of the object from its equilibrium position:
x ≈ (0.5 * 14.4 N) / (39.4 N/m * 0.9446)
x ≈ (7.2 N) / (37.22184 N/m)
x ≈ 0.1934 m
So, the displacement is approximately 0.1934 meters from the equilibrium position.
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