A uniform beam of mass M = 10.0 kg and length L = 2.00 m is suspended from a ceiling at an angle of 30° below the horizontal. The beam is attached to the ceiling by a hinge at one end and a light support rope attached to the other end. What is the magnitude of the force by the hinge? 30° L= 2.0 m M= 10.0 kg

A uniform beam of mass M = 10.0 kg and length L = 2.00 m is suspended from a ceiling at an angle of 30° below the horizontal. The beam is attached to the ceiling by a hinge at one end and a light support rope attached to the other end. What is the magnitude of the force by the hinge? 30° L= 2.0 m M= 10.0 kg

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter12: Static Equilibrium And Elasticity

Section: Chapter Questions

Problem 30AP

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In Example 14.3, we found that one of the steel cables supporting an airplane at the Udvar-Hazy Center was under a tension of 9.30 103 N. Assume the cable has a diameter of 2.30 era and an initial length of 8.00 m before the plane is suspended on the cable. How much longer is the cable when the plane is suspended on it?
A wooden door 2.1 m high and 0.90 m wide is hung by two hinges 1.8 m apart. The lower hinge is 15 cm above the bottom of the door. The center of mass of the door is at its geometric center, and the weight of the door is 260 N, which is supported equally by both hinges. Find the horizontal force exerted by each hinge on the door.
A bridge of length 50.0 m and mass 8.00 104 kg is supported on a smooth pier at each end as shown in Figure P12.25. A truck of mass 3.00 104 kg is located 15.0 m from one end. What are the forces on the bridge at the points of support? Figure P12.25
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Ruby, with mass 55.0 kg, is trying to reach a box on a high shelf by standing on her tiptoes. In this position, half her weight is supported by the normal force exerted by the floor on the toes of each foot as shown in Figure P14.75A. This situation can be modeled mechanically by representing the force on Rubys Achilles tendon with FA and the force on her tibia as FT as shown in Figure P14.75B. What is the value of the angle and the magnitudes of the forces FA and FT? FIGURE P14.75
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A 10.0-kg monkey climbs a uniform ladder with weight 1.20 102 N and length L = 3.00 m as shown in Figure P12.14. The ladder rests against the wall and makes an angle of = 60.0 with the ground. The upper and lower ends of the ladder rest on frictionless surfaces. The lower end is connected to the wall by a horizontal rope that is frayed and can support a maximum tension of only 80.0 N. (a) Draw a force diagram for the ladder. (b) Find the normal force exerted on the bottom of the ladder. (c) Find the tension in the rope when the monkey is two-thirds of the way up the ladder. (d) Find the maximum distance d that the monkey can climb up the ladder before the rope breaks. (e) If the horizontal surface were rough and the rope were removed, how would your analysis of the problem change? What other information would you need to answer parts (c) and (d)? Figure P12.14
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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