Physics for Scientists and Engineers, Volume 2
10th Edition
ISBN: 9781337553582
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 12, Problem 13P
Figure P12.13 shows a claw hammer being used to pull a nail out of a horizontal board. The mass of the hammer is 1.00 kg. A force of 150 N is exerted horizontally as shown, and the nail does not yet move relative to the board. Find (a) the force exerted by the hammer claws on the nail and (b) the force exerted by the surface on the point of contact with the hammer head. Assume the force the hammer exerts on the nail is parallel to the nail.
Figure P12.13
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In exercise physiology it is sometimes important to determine the location of a person's center of gravity with respect to the plantar area. This can be done with the arrangement shown in the figure. A lightweight plank is supported by two scales that indicate Fg1 = 380.0 N and Fg2 = 320.0 N. The scales are separated by a distance of 2.00 m. At what distance from the woman's feet is her center of gravity? If a) the weight of the plank is neglected. b) if the plank has a weight of 55.0 N.
Chapter 12 Solutions
Physics for Scientists and Engineers, Volume 2
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 - You are building additional storage space in your...Ch. 12 - Why is the following situation impossible? A...Ch. 12 - Prob. 3PCh. 12 - A circular pizza of radius R has a circular piece...Ch. 12 - Your brother is opening a skateboard shop. He has...Ch. 12 - A uniform beam of length 7.60 m and weight 4.50 ...
Ch. 12 - Prob. 7PCh. 12 - A uniform beam of length L and mass m shown in...Ch. 12 - A flexible chain weighing 40.0 N hangs between two...Ch. 12 - A 20.0-kg floodlight in a park is supported at the...Ch. 12 - Prob. 11PCh. 12 - Review. While Lost-a-Lot ponders his next move in...Ch. 12 - Figure P12.13 shows a claw hammer being used to...Ch. 12 - A 10.0-kg monkey climbs a uniform ladder with...Ch. 12 - John is pushing his daughter Rachel in a...Ch. 12 - Prob. 16PCh. 12 - The deepest point in the ocean is in the Mariana...Ch. 12 - A steel wire of diameter 1 mm can support a...Ch. 12 - A child slides across a floor in a pair of...Ch. 12 - Evaluate Youngs modulus for the material whose...Ch. 12 - Assume if the shear stress in steel exceeds about...Ch. 12 - When water freezes, it expands by about 9.00%....Ch. 12 - Review. A 30.0-kg hammer, moving with speed 20.0...Ch. 12 - A uniform beam resting on two pivots has a length...Ch. 12 - A bridge of length 50.0 m and mass 8.00 104 kg is...Ch. 12 - Prob. 26APCh. 12 - The lintel of prestressed reinforced concrete in...Ch. 12 - The following equations are obtained from a force...Ch. 12 - A hungry bear weighing 700 N walks out on a beam...Ch. 12 - A 1 200-N uniform boom at = 65 to the vertical is...Ch. 12 - A uniform sign of weight Fg and width 2L hangs...Ch. 12 - When a person stands on tiptoe on one foot (a...Ch. 12 - A 10 000-N shark is supported by a rope attached...Ch. 12 - Assume a person bends forward to lift a load with...Ch. 12 - A uniform beam of mass m is inclined at an angle ...Ch. 12 - Why is the following situation impossible? A...Ch. 12 - When a circus performer performing on the rings...Ch. 12 - Figure P12.38 shows a light truss formed from...Ch. 12 - Prob. 39APCh. 12 - A stepladder of negligible weight is constructed...Ch. 12 - A stepladder of negligible weight is constructed...Ch. 12 - Review. A wire of length L, Youngs modulus Y, and...Ch. 12 - Two racquetballs, each having a mass of 170 g, are...Ch. 12 - Prob. 44APCh. 12 - Review. An aluminum wire is 0.850 m long and has a...Ch. 12 - You have been hired as an expert witness in a case...Ch. 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 - A uniform rod of weight Fg and length L is...Ch. 12 - In the What If? section of Example 12.2, let d...
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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 P14.75arrow_forwardWhy 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 stepladder of negligible weight is constructed as shown in Figure P10.73, with AC = BC = ℓ. A painter of mass m stands on the ladder a distance d from the bottom. Assuming the floor is frictionless, find (a) the tension in the horizontal bar DE connecting the two halves of the ladder, (b) the normal forces at A and B, and (c) the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half. Suggestion: Treat the ladder as a single object, but also treat each half of the ladder separately. Figure P10.73 Problems 73 and 74.arrow_forward
- When 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 beam resting on two pivots has a length L = 6.00 m and mass M = 90.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot located a distance = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 55.0 kg steps onto the left end of the beam and begins walking to the right as in Figure P10.28. The goal is to find the womans position when the beam begins to tip. (a) What is the appropriate analysis model for the beam before it begins to tip? (b) Sketch a force diagram for the beam, labeling the gravitational and normal forces acting on the beam and placing the woman a distance x to the right of the first pivot, which is the origin. (c) Where is the woman when the normal force n1 is the greatest? (d) What is n1 when the beam is about to tip? (e) Use Equation 10.27 to find the value of n2 when the beam is about to tip. (f) Using the result of part (d) and Equation 10.28, with torques computed around the second pivot, find the womans position x when the beam is about to tip. (g) Check the answer to part (e) by computing torques around the first pivot point. Figure P10.28arrow_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_forward
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- The fishing pole in Figure P10.22 makes an angle of 20.0 with the horizontal. What is the torque exerted by the fish about an axis perpendicular to the page and passing through the anglers hand if the fish pulls on the fishing line with a force F=100N at an angle 37.0 below the horizontal? The force is applied at a point 2.00 m from the anglers hands. Figure P10.22arrow_forwardWhy is the following situation impossible? A worker in a factory pulls a cabinet across the floor using a rope as shown in Figure P12.36a. The rope make an angle = 37.0 with the floor and is tied h1 = 10.0 cm from the bottom of the cabinet. The uniform rectangular cabinet has height = 100 cm and width w = 60.0 cm, and it weighs 400 N. The cabinet slides with constant speed when a force F = 300 N is applied through the rope. The worker tires of walking backward. He fastens the rope to a point on the cabinet h2 = 65.0 cm off the floor and lays the rope over his shoulder so that he can walk forward and pull as shown in Figure P12.36b. In this way, the rope again makes an angle of = 37.0 with the horizontal and again has a tension of 300 N. Using this technique, the worker is able to slide the cabinet over a long distance on the floor without tiring. Figure P12.36 Problems 36 and 44.arrow_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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