Essential University Physics: Volume 1; Mastering Physics with Pearson eText -- ValuePack Access Card -- for Essential University Physics (3rd Edition)
3rd Edition
ISBN: 9780134197319
Author: Richard Wolfson
Publisher: PEARSON
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Chapter 12, Problem 20E
A 4.2-m-long beam is supported by a cable at its center. A 65-kg steelworker stands at one end of the beam. Where should a 190-kg bucket of concrete be suspended for the beam to be in static equilibrium?
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A 10 m long uniform board weighing 400 N is supported by ropes at each end. A worker weighing 900 N stands 3.0 m from the rope on the left. Determine the tension in the two ropes for static equilibrium.
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Chapter 12 Solutions
Essential University Physics: Volume 1; Mastering Physics with Pearson eText -- ValuePack Access Card -- for Essential University Physics (3rd Edition)
Ch. 12.1 - The figure shows three pairs of forces acting on...Ch. 12.2 - Prob. 12.2GICh. 12.3 - The figure shows a person in static equilibrium...Ch. 12.4 - Prob. 12.4GICh. 12 - Give an example of an object on which the net...Ch. 12 - Give an example of an object on which the net...Ch. 12 - Prob. 3FTDCh. 12 - Pregnant women often assume a posture with their...Ch. 12 - When you carry a bucket of water with one hand,...Ch. 12 - Is a ladder more likely to slip when you stand...
Ch. 12 - How does a heavy keel help keep a boat from...Ch. 12 - Does choosing a pivot point in an equilibrium...Ch. 12 - If you take the pivot point at the application...Ch. 12 - A short dog and a tall person are standing on a...Ch. 12 - Prob. 11FTDCh. 12 - A body is subject to three forces; F1=1i+2jN,...Ch. 12 - To demonstrate that the choice of pivot point...Ch. 12 - In Fig. 12.11 the forces shown all have the same...Ch. 12 - Figure 12.12a shows a thin, uniform square plate...Ch. 12 - Repeat the preceding problem for the equilateral...Ch. 12 - A 23-m-long log of irregular cross section lies...Ch. 12 - A 60-kg uniform board 2.4 m long is supported by a...Ch. 12 - Where should the child in Fig. 12.14 sit if the...Ch. 12 - A 4.2-m-long beam is supported by a cable at its...Ch. 12 - Figure 12.15 shows how a scale with a capacity of...Ch. 12 - A portion of a roller-coaster track is described...Ch. 12 - Prob. 23ECh. 12 - Youre a highway safety engineer, and youre asked...Ch. 12 - Figure 12.17a shows an outstretched arm with mass...Ch. 12 - A uniform sphere of radius R is supported by a...Ch. 12 - You work for a garden equipment company, and youre...Ch. 12 - Figure 12.20 shows the fool and lower leg of a...Ch. 12 - A uniform 5.0-kg ladder is leaning against a...Ch. 12 - The boom in the crane of Fig. 12.21 is free to...Ch. 12 - A uniform board of length L and weight W is...Ch. 12 - Figure 12.23 shows a 1250-kg car that has slipped...Ch. 12 - Repeat Example 12.2, now assuming that the...Ch. 12 - You are headwaiter at a new restaurant, and your...Ch. 12 - Climbers attempting to cross a stream place a...Ch. 12 - A crane in a marble quarry is mounted on the...Ch. 12 - A rectangular block measures w w L, where L is...Ch. 12 - The potential energy as a function of position for...Ch. 12 - A rectangular block of mass m measures w w L,...Ch. 12 - A 160-kg highway sign of uniform density is 2.3 m...Ch. 12 - A 5.0-m-long ladder has mass 9.5 kg and is leaning...Ch. 12 - Prob. 42PCh. 12 - A uniform, solid cube of mass m and side s is in...Ch. 12 - An isosceles triangular block of mass m and height...Ch. 12 - Youre investigating ladder safety for the Consumer...Ch. 12 - A 2.0-m-long rod has density in kilograms per...Ch. 12 - What horizontal force applied at its highest point...Ch. 12 - A rectangular block twice as high as it is wide is...Ch. 12 - What condition on the coefficient of friction in...Ch. 12 - A uniform solid cone of height h and base diameter...Ch. 12 - Prove the statement in Section 12.1 that the...Ch. 12 - Three identical books of length L are stacked over...Ch. 12 - A uniform pole of mass M is at rest on an incline...Ch. 12 - For what angle does the situation in Problem 53...Ch. 12 - Figure 12.31 shows a popular system for mounting...Ch. 12 - The nuchal ligament is a thick, cordlike structure...Ch. 12 - A 4.2-kg plant hangs from the bracket shown in...Ch. 12 - The wheel in Fig. 12.34 has mass M and is weighted...Ch. 12 - An interstellar spacecraft from an advanced...Ch. 12 - Youre called to testify in a product liability...Ch. 12 - Youre designing a vacation cabin at a ski resort....Ch. 12 - Prob. 62PCh. 12 - Engineers designing a new semiconductor device...Ch. 12 - Youve been hired by your states environmental...Ch. 12 - Youve been hired by your states environmental...Ch. 12 - Youve been hired by your states environmental...Ch. 12 - Youve been hired by your states environmental...
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- 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.arrow_forwardProblems 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_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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