Problem 1: The figure presented here shows a horizontal beam of uniform density in static equilibrium. The beam has a frictionless support at 1/4 from the left edge of the beam, and total beam length of the L. A force, F, acts at the end of the beam in the +y direction. The mass of the beam per-unit-length is mx. The acceleration due to gravity is g. L/4 Part A: What is the total force due the weight of the beam and where does this force act in the system? Part B: Draw the FBD of the beam. Part C: Find an equation for Fsupport (the support reaction force). Express your answer in terms of F (the applied force at the end of the beam), mx, L and g.
Problem 1: The figure presented here shows a horizontal beam of uniform density in static equilibrium. The beam has a frictionless support at 1/4 from the left edge of the beam, and total beam length of the L. A force, F, acts at the end of the beam in the +y direction. The mass of the beam per-unit-length is mx. The acceleration due to gravity is g. L/4 Part A: What is the total force due the weight of the beam and where does this force act in the system? Part B: Draw the FBD of the beam. Part C: Find an equation for Fsupport (the support reaction force). Express your answer in terms of F (the applied force at the end of the beam), mx, L and g.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Solve handwriting Solutions all parts. I req to u ok .........
Transcribed Image Text:Problem 1:
The figure presented here shows a horizontal beam of uniform density in static
equilibrium. The beam has a frictionless support at 1/4 from the left edge of the
beam, and total beam length of the L. A force, F, acts at the end of the beam in the
+y direction. The mass of the beam per-unit-length is mx. The acceleration due to
gravity is g.
L
Part A: What is the total force due the weight of the beam and where does this
force act in the system?
Part B: Draw the FBD of the beam.
Part C: Find an equation for Fsupport (the support reaction force). Express your
answer in terms of F (the applied force at the end of the beam), mx, L and g.
L/4
F
L
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