Sample Problem 12.02 Balancing a leaning boomFigure 12-6a shows a safe (mass M = 430 kg) hanging by arope (negligible mass) from a boom (a = 1.9 m and b =2.5 m) that consists of a uniform hinged beam (m = 85 kg)and horizontal cable (negligible mass).Figure 12-6 (a)A heavy safe ishung from aboom consistingCableof a horizontal(a) What is the tension T, in the cable? In other words, what isthe magnitude of the force T on the beam from the cable?steel cable and auniform beam.Beam comRope(b) A free-bodydiagram for thebeam.KEY IDEASThe system here is the beam alone, and the forces on it areshown in the free-body diagram of Fig. 12-6b. The forcefrom the cable is T. The gravitational force on the beamacts at the beam's center of mass (at the beam's center) andis represented by its equivalent mg. The vertical componentof the force on the beam from the hinge is F, and the hori-zontal component of the force from the hinge is F. Theforce from the rope supporting the safe is T. Because beam,rope, and safe are stationary, the magnitude of T, is equal tothe weight of the safe: T, = Mg. We place the origin O of anxy coordinate system at the hinge. Because the system is instatic equilibrium, the balancing equations apply to it.Hinge(a)MBeammgHere is theF,wise choice ofCalculations: Let us start with Eq. 12-9 (Tnet.z = 0). Notethat we are asked for the magnitude of force T, and not offorces F, and F, acting at the hinge, at point O. To eliminateF, and F, from the torque calculation, we should calculatetorques about an axis that is perpendicular to the figure atpoint O. Then F, and F will have moment arms of zero. Thelines of action for T, T, and mg are dashed in Fig. 12-6b.The corresponding moment arms are a, b, and b/2.Writing torques in the form of r Fand using our rule aboutsigns for torques, the balancing equation Tet.z = 0 becomesrotation axis.(b)Fmagnitude F of the net force. Because we know T, we applythe force balancing equations to the beam.Calculations: For the horizontal balance, we can rewriteFst = 0 asF, – T = 0,(12-20)(a)(T.) – (b)(T,) – (¿b)(mg) = 0.(12-19)and soF, = T = 6093 N.Substituting Mg for T, and solving for T, we find thatFor the vertical balance, we write Fety = 0 asgb(M + m)F, - mg – T, = 0.T =aSubstituting Mg for T, and solving for F, we find that(9.8 m/s²)(2.5 m)(430 kg + 85/2 kg)F, = (m + M)g = (85 kg + 430 kg)(9.8 m/s²)1.9 m= 6093 N = 6100 N.(Answer)= 5047 N.From the Pythagorean theorem, we now have(b) Find the magnitude F of the net force on the beam fromthe hinge.F = VF} + F2= V(6093 N)? + (5047 N)² = 7900 N.(Answer)KEY IDEANow we want the horizontal component F, and verticalcomponent F, so that we can combine them to get theNote that Fis substantially greater than either the combinedweights of the safe and the beam, 5000 N, or the tension inthe horizontal wire, 6100 N.PLUS Additional examples, video, and practice available at WileyPLUS In sample problem 12.02, the vertical forceacting on the beam by the hinge is equal toа. Mgb. Tr+ mgС.Trd. Tc

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Answered: In sample problem 12.02, the vertical…

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In sample problem 12.02, the vertical force acting on the beam by the hinge is equal to а. Mg b. Tr+ mg С. Tr d. To

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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