When the structure shown below is supported at point P, it is in equilibrium. Find the magnitude of force F and the force applied at P. The weight of the structure isnegligible.2.3 mF4.6 m2.3 m-2.3 m1800 NР3.5 m800 N900 NSubmit Answer Incorrect. Tries 2/40 Previous TriesFind the force applied at P.Magnitude =Direction =Submit Answer Tries 0/40(wrt the +x-axis)K

Draw the free-body diagram of the system.

When the structure shown below is supported at point P, it is in equilibrium. Find the magnitude of force F and the force applied at P. The weight of the structure is negligible. 2.3 F 4.6 m2.3 m 2.3 m 1800 N P 3.5 m 1600 N 900 N Submit Answer Incorrect. Tries 2/40 Previous Tries Find the force applied at P. Magnitude = Submit Answer Tries 0/40 Direction = (wrt the +x-axis) K

When the structure shown below is supported at point P, it is in equilibrium. Find the magnitude of force F and the force applied at P. The weight of the structure is negligible. 2.3 F 4.6 m2.3 m 2.3 m 1800 N P 3.5 m 1600 N 900 N Submit Answer Incorrect. Tries 2/40 Previous Tries Find the force applied at P. Magnitude = Submit Answer Tries 0/40 Direction = (wrt the +x-axis) K

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter14: Static Equilibrium, Elasticity, And Fracture

Section: Chapter Questions

Problem 33PQ: Problems 33 and 34 are paired. One end of a uniform beam that weighs 2.80 102 N is attached to a...

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At a museum, a 1300-kg model aircraft is hung from a lightweight beam of length 12.0 m that is free to pivot about its base and is supported by a massless cable (Fig. P14.38). Ignore the mass of the beam. a. What is the tension in the section of the cable between the beam and the wall? b. What are the horizontal and vertical forces that the pivot exerts on the beam? FIGURE P14.38 (a) From the free-body diagram, the angle that the string tension makes with the beam is = 55.0 + 18.0 = 73.0, and the perpendicular component of the string tension is FT sin73.0. Summing torques around the base of the rod gives (Eq. 14.2): =0:(12.0m)(1300kg)(9.81m/s2)cos55.0+FT(12.0m)sin73.0=0FT=(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0FT=7.65103N Figure P14.38ANS (b) Using force balance (Eq. 14.1): Fx=0:FHFTcos18.0=0FH=FTcos18.0=[(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0]cos18.0=7.27103NFy=0:FVFTsin18.0(1300kg)(9.81m/s2)=0 FV=FTsin18.0+(1300kg)gFV=[(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0]sin18.0+(1300kg)(9.81m/s2)FV=1.51104N
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