Consider a rigid 2.20-m-long beam (see figure) that is supported by a fixed 1.10

Question

Consider a rigid 2.20-m-long beam (see figure) that is supported by a fixed 1.10-m-high post through its center and pivots on a frictionless bearing at its center atop the vertical 1.10-m-high post. One end of the beam is connected to the floor by a spring that has a force constant k = 1500 N/m. when the beam is horizontal, the spring is vertical and unstressed. If an object is hung from the opposite end of the beam, the beam settles into an equilibrium position where it makes an angle of 22.1° with the horizontal. What is the mass of the object? connected to the foo b ttles into an equlibrium x kg Enter a number.

Explanation / Answer

Solution:

Weight of the unknown mass = mg

Spring constant = k = 1500 N/m

Torque die to the mass m about the center = (mg cos 22.1 )(1.10) = 9.988 m Nm

torque due to the extension of spring = (-kx) cos 22.1 * 1.10 = -1500 (1.1 sin22.1) cos22.1 * 1.1 = 632.7 Nm

9.988 m = 632.7

=> Mass = m = 632.7 / 9.988 = 63.3 kg

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