Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 12, Problem 12P
(a)
To determine
The magnitudes of
(b)
To determine
The magnitudes of
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Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Ch. 12.1 - Consider the object subject to the two forces of...Ch. 12.1 - Consider the object subject to the three forces in...Ch. 12.2 - A meterstick of uniform density is hung from a...Ch. 12.4 - For the three parts of this Quick Quiz, choose...Ch. 12 - Prob. 1OQCh. 12 - Prob. 2OQCh. 12 - Prob. 3OQCh. 12 - Prob. 4OQCh. 12 - In the cabin of a ship, a soda can rests in a...Ch. 12 - Prob. 6OQ
Ch. 12 - Prob. 7OQCh. 12 - Prob. 8OQCh. 12 - Prob. 9OQCh. 12 - Prob. 10OQCh. 12 - Prob. 1CQCh. 12 - Prob. 2CQCh. 12 - Prob. 3CQCh. 12 - Prob. 4CQCh. 12 - Prob. 5CQCh. 12 - Prob. 6CQCh. 12 - Prob. 7CQCh. 12 - What kind of deformation does a cube of Jell-O...Ch. 12 - Prob. 1PCh. 12 - Why is the following situation impossible? A...Ch. 12 - Prob. 3PCh. 12 - Prob. 4PCh. 12 - Your brother is opening a skateboard shop. He has...Ch. 12 - A circular pizza of radius R has a circular piece...Ch. 12 - Prob. 7PCh. 12 - Prob. 8PCh. 12 - Prob. 9PCh. 12 - Prob. 10PCh. 12 - A uniform beam of length 7.60 m and weight 4.50 ...Ch. 12 - Prob. 12PCh. 12 - Prob. 13PCh. 12 - A uniform ladder of length L and mass m1 rests...Ch. 12 - A flexible chain weighing 40.0 N hangs between two...Ch. 12 - A uniform beam of length L and mass m shown in...Ch. 12 - Figure P12.13 shows a claw hammer being used to...Ch. 12 - A 20.0-kg floodlight in a park is supported at the...Ch. 12 - Prob. 19PCh. 12 - Review. While Lost-a-Lot ponders his next move in...Ch. 12 - John is pushing his daughter Rachel in a...Ch. 12 - Prob. 22PCh. 12 - Prob. 23PCh. 12 - A 10.0-kg monkey climbs a uniform ladder with...Ch. 12 - Prob. 25PCh. 12 - A steel wire of diameter 1 mm can support a...Ch. 12 - The deepest point in the ocean is in the Mariana...Ch. 12 - Assume Youngs modulus for bone is 1.50 1010 N/m2....Ch. 12 - A child slides across a floor in a pair of...Ch. 12 - Evaluate Youngs modulus for the material whose...Ch. 12 - Prob. 31PCh. 12 - When water freezes, it expands by about 9.00%....Ch. 12 - Prob. 33PCh. 12 - Prob. 34PCh. 12 - Prob. 35PCh. 12 - Review. A 30.0-kg hammer, moving with speed 20.0...Ch. 12 - A bridge of length 50.0 m and mass 8.00 104 kg is...Ch. 12 - A uniform beam resting on two pivots has a length...Ch. 12 - Prob. 39APCh. 12 - The lintel of prestressed reinforced concrete in...Ch. 12 - Prob. 41APCh. 12 - When a person stands on tiptoe on one foot (a...Ch. 12 - A hungry bear weighing 700 N walks out on a beam...Ch. 12 - Prob. 44APCh. 12 - A uniform sign of weight Fg and width 2L hangs...Ch. 12 - Prob. 46APCh. 12 - Prob. 47APCh. 12 - Assume a person bends forward to lift a load with...Ch. 12 - A 10 000-N shark is supported by a rope attached...Ch. 12 - Prob. 50APCh. 12 - A uniform beam of mass m is inclined at an angle ...Ch. 12 - Prob. 52APCh. 12 - When a circus performer performing on the rings...Ch. 12 - Figure P12.38 shows a light truss formed from...Ch. 12 - Prob. 55APCh. 12 - A stepladder of negligible weight is constructed...Ch. 12 - A stepladder of negligible weight is constructed...Ch. 12 - Prob. 58APCh. 12 - Two racquetballs, each having a mass of 170 g, are...Ch. 12 - Review. A wire of length L, Youngs modulus Y, and...Ch. 12 - Review. An aluminum wire is 0.850 m long and has a...Ch. 12 - Prob. 62APCh. 12 - A 500-N uniform rectangular sign 4.00 m wide and...Ch. 12 - A steel cable 3.00 cm2 in cross-sectional area has...Ch. 12 - Prob. 65CPCh. 12 - In the What If? section of Example 12.2, let d...Ch. 12 - Prob. 67CPCh. 12 - A uniform rod of weight Fg and length L is...
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- Why is the following situation impossible? A uniform beam of mass mk = 3.00 kg and length = 1.00 m supports blocks with masses m1 = 5.00 kg and m2 = 15.0 kg at two positions as shown in Figure P12.2. The beam rests on two triangular blocks, with point P a distance d = 0.300 m to the right of the center of gravity of the beam. The position of the object of mass m2 is adjusted along the length of the beam until the normal force on the beam at O is zero. Figure P12.2arrow_forwardA 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 P14.78arrow_forwardRuby, 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 P14.75arrow_forward
- A uniform beam of length 7.60 m and weight 4.50 102 N is carried by two workers, Sam and Joe, as shown in Figure P12.6. Determine the force that each person exerts on the beam. Figure P12.6arrow_forwardWhen a circus performer performing on the rings executes the iron cross, he maintains the position at rest shown in Figure P12.37a. In this maneuver, the gymnasts feet (not shown) are off the floor. The primary muscles involved in supporting this position are the latissimus dorsi (lats) and the pectoralis major (pecs). One of the rings exerts an upward force Fk on a hand as show n in Figure P12.37b. The force Fs, is exerted by the shoulder joint on the arm. The latissimus dorsi and pectoralis major muscles exert a total force Fm on the arm. (a) Using the information in the figure, find the magnitude of the force Fm for an athlete of weight 750 N. (b) Suppose a performer in training cannot perform the iron cross but can hold a position similar to the figure in which the arms make a 45 angle with the horizontal rather than being horizontal. Why is this position easier for the performer? Figure P12.37arrow_forwardA uniform sign of weight Fg and width 2L hangs from a light, horizontal beam hinged at the wall and supported by a cable (Fig. P12.31). Determine (a) the tension in the cable and (b) the components of the reaction force exerted by the wall on the beam in terms of Fg, d, L, and . Figure P12.31arrow_forward
- 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 P12.14arrow_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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