27. What force must be applied at point P to keep the structure shown in equilibrium? The weight of the structure is negligible. P 2.0 m 2.0 m - 4000 N 1.0 m 2000 N 28. Is it possible to apply a force at P to keep in equilibrium the structure shown? The weight of the structure is negligible. 2000 N 2.0 m P 2.0 m 5000 N 1.0 m 3000 N

27. What force must be applied at point P to keep the structure shown in equilibrium? The weight of the structure is negligible. P 2.0 m 2.0 m - 4000 N 1.0 m 2000 N 28. Is it possible to apply a force at P to keep in equilibrium the structure shown? The weight of the structure is negligible. 2000 N 2.0 m P 2.0 m 5000 N 1.0 m 3000 N

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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Assume Youngs modulus for bone is 1.50 1010 N/m2. The bone breaks if stress greater than 1.50 108 N/m2 is imposed on it. (a) What is the maximum force that can be exerted on the femur bone in the leg if it has a minimum effective diameter of 2.50 cm? (b) If this much force is applied compressively, by how much does the 25.0-cm-long bone shorten?
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
A massless, horizontal beam of length L and a massless rope support a sign of mass m (Fig. P14.78). a. What is the tension in the rope? b. In terms of m, g, d, L, and , what are the components of the force exerted by the beam on the wall? FIGURE P14.78
What Is Static Equilibrium? Problems 13 are grouped. 1. C A ball is attached to a strong, lightweight rod (Fig. P14.1). The rod is supported by a horizontal pin near the top. The ball is at rest. Is the ball in static equilibrium? If not, why not? If so, which type of equilibrium is itstable, unstable, or neutral? Hint: What would happen if you displaced the ball slightly? FIGURE P14.1
A rod, 120cm long and of diameter 3.0 cm is stretched by a force of 18 kN. What is its stress? O a. 25.47 N/mm² O b. 20.70 N/mm² O c. 2,470 N/mm² O d. 245 N/mm²
A 100 m long cable with a cross-sectional area of 1.0 mm² stretches by 5.0 cm when a force of 1,000 N is applied. The Young's Modulus is Select one: Oa. 1.0 x 10¹3 N/m². O b. 5.0 x 10¹¹ N/m² O c. 5.0 x 10¹3 N/m². O d. None of these choices. 1.0 x 10¹¹ N/m². 2.0 x 10¹2 N/m². e. Of.
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A rod 120 cm long and of a diameter 3.0 cm is stretched by a force of 18 kN. What is its stress? a. 20. 70 N/mm^2 b. 25.47 N/mm^2 c. 245 N/mm^2 d. 2,470 N/mm^2
1. please answer ASAP
A cable used to lift heavy materials like steel I-beams must be strong enough to resist breaking even under a load of 1.0 x 10^6 N. For safety, the cable must support twice that load. a. What cross-sectional area should the cable have if it's to be made of steel? b. By how much will an 8.0-m length of this cable stretch when subject to the 1.0 x 10^6 N load?Tensile strength of steel = 5.0 x 10^8Answers a. 4.0 x 10^-3 m^2 b. 1.0 x 10^-2 m
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 m F 4.6 m2.3 m -2.3 m 1800 N Р 3.5 m 800 N 900 N Submit Answer Incorrect. Tries 2/40 Previous Tries Find the force applied at P. Magnitude = Direction = Submit Answer Tries 0/40 (wrt the +x-axis) K
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