A beam rests upon two springs. When
force F=F= 700 NN is applied to the beams, the
springs compress and the beam becomes perfectly horizontal.
A Beam Rests Upon Two Springs When Force F F 700 Nn Is Applied To The Beams The Springs Compress And The Beam Becomes 1 (172.95 KiB) Viewed 30 times
A beam rests upon two springs. When force F = 700 N is applied to the beams, the springs compress and the beam becomes perfectly horizontal. F L1 L2 B WWW Geometry: L = 0.6 m L = 1.9 m Spring Constants: ka = 3.9 kN/m kB = 6 kN/m i. Determine the magnitude of force acting on the beam from spring A. ii. Determine the magnitude of force acting on the beam from spring B. iii. Determine is the deflection of spring A. (i.e. what is the distance that spring A is squished?) iv. Determine is the deflection of spring B.(i.e. what is the distance that spring B is squished?) v. Force F is then removed and the springs decompress and return to their resting position. The beam is no longer horizontal. Determine the angle of tilt 0 of the beam from the horizontal. Assumptions you can make: • The beam is massless and rigid. • The springs can only move in the vertical direction.
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