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During a walk on a rope, a tightrope walker creates a tension of
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- A steel cable 2.00 m in length and with cross-sectional radius 0.350 mm is used to suspend from the ceiling a 10.0-kg model aircraft that is flying in a horizontal circle with an angular speed of 6.00 rad/s. What is the strain produced in the cable?arrow_forwardIn 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?arrow_forwardAn aluminium (=2.7g/cm3) wire is suspended from the ceiling and hangs vertically. How long must the wire be before the stress at its upper end reaches the proportionality limit, which is 8.0107N/m2 ?arrow_forward
- Two rods, one made of copper and the other of steel, have the same dimensions. If the copper rod stretches by 0.15mm under some stress, how much does the steel rod stretch under the same stress?arrow_forwardA 100-N weight is attached to a free end of a metallic wire that hangs from the ceiling. When a second 100-N weight is added to the wire, it stretches 3.0 mm. The diameter and the length of the wire are 1.0 mm and2.0 m, respectively. What is Young’s modulus of the metal used to manufacture the wire?arrow_forwardWhat 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 P14.1arrow_forward
- A 5.45-N beam of uniform density is 1.60 m long. The beam is supported at an angle of 35.0 by a cable attached to one end. There is a pin through the other end of the beam (Fig. P14.30). Use the values given in the figure to find the tension in the cable. FIGURE P14.30arrow_forwardConsider the sketch of a portion of a roller-coaster track seen in Figure P14.5. Identify places on the track that could be considered possible locations of static equilibrium for a rollercoaster car were the car to be placed at any spot on the track.Which places are candidate locations for stable, unstable, andneutral static equilibrium? FIGURE P14.5arrow_forwardAssume 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?arrow_forward
- Problems 33 and 34 are paired. One end of a uniform beam that weighs 2.80 102 N is attached to a wall with a hinge pin. The other end is supported by a cable making the angles shown in Figure P14.33. Find the tension in the cable. FIGURE P14.33 Problems 33 and 34.arrow_forwardA flexible chain weighing 40.0 N hangs between two hooks located at the same height (Fig. P12.9). At each hook, the tangent to the chain makes an angle = 42.0 with the horizontal. Find (a) the magnitude of the force each hook exerts on the chain and (b) the tension in the chain at its midpoint. Suggestion: For part (b), make a force diagram for half of the chain. Figure P12.9arrow_forwardAt 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=1.51104Narrow_forward
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