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S 0% D Wed 7:37 PM a E Firefox File Edit View History Bookmarks Tools Window Help Physics 150 2015, Chapter 8, Assig VA Physics 150 2015, Chapter 3 a search webassig n.net/web/Student/Assignment-Responses/submit?dep 12312706 Submit Answer Save Progress 3. 0/3 points Previous Answers My Notes Ask Your Teacher A lamp of mass mL 19.0 kg hangs at the end of a rod of mass mR 12.0 kg and length L 1.60 m. The left end of the rod is attached to a hinged bracket fixed in a wall. A cable, also anchored to the wall, is connected to the rod at a point a distance d 1.10 m from the wall. The cable makes an angle 6 51.0 as defined in the drawing, with the rod Begin by drawing an extended free body diagram for the bar. Remember that since you do not know the magnitude o angle of the force on the hinge, you should draw the x and y components of this force separately. Also, you will assume that the rod is "uniform" and that you may think of its mass as being concentrated at the middle of the rod Retain at least four significant figures in all calculated results that will be used in further calculations. Round only the submitted answers to three significant figures.) Lectu Calculate the tension in the cable. H NT It is easiest to use the hinge as the rotational axis when you apply the condition of rotational equilibrium. +Bio31 298.56 N Calculate the horizontal component of the support force applied by the wall hinge) to the bar. HINT: The easiest method here is to look at the condition of static equilibrium by considering the sum of the forces in the x-direction wall hinge) to the bar. HINT: The easiest method here is to look at the condition of static equilibrium by Calculate the vertical component of the support force applied by the considering the sum of the forces in the y-direction Submit Assignment Save Assignment Progress Home My Assignments Extension Request WebAssign 4.0 1997-2015 Advanced Instruct nc. All rightExplanation / Answer
Here ,
let the tension is T
balancing the moment of forces about the hinge
T * sin(51) * d - mg * L - mR * g * L/2 = 0
T * sin(51) * 1.1 - 12 * 9.8 * (1.6/2) - 19 * 9.8 * 1.6 = 0
sovlving for T
T = 458.6 N
the tension in the cable is 458.6 N
in horizontal direction
Fh = T * cos(theta)
Fh = 458.6 * cos(51)
Fh = 288.6 N
in vertical direction
Fv - 19 * g - 12 * g + 458.6 * sin(51) = 0
solving for Fv
Fv = -52.3 N
the vertical force at the support hinge is -52.3 N
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