, Fall17-123-Static Equ © A person Working At A [ Course Documents-H x D example
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, Fall17-123-Static Equ © A person Working At A [ Course Documents-H x D example test2 phys a12 Cooper x X Secure | https://www.webassign.net/web/Student/Assignment-Responses/submit2deps16554943 6. 0/1 points | Previous Answers My Notes As shown in the figure below, a uniform beam is supported by a cable at one end and the force of friction at the other end. The cable makes an angle of : 30 telength ofthe beam is L = 3.75 m, the coefficient of static friction between the wall and the beam is 4-0.440, and the weight of the beam is represented by w. Determine the minimum distance x from x = 1.135 point A at which an additional wight 2w (twice the weight of the rod) can be hung without causing the rod to slip at point A Since this is an equilibrium problem, we expect to use the first and second conditions of equilibrium to solve the problem. Before starting the problem, a well-drawn figure with all forces and all forces resolved into components would be a real asset to finding a solution. See if you can use the first condition of equilibrium to obtain an expression for the normal force and then the force of friction. Next see if you can use the first condition of equilibrium to obtain an expression for the tension in the cable. Finally see if you can use the second condition of equilibrium to obtain an expression that will allow you to determine the maximum distance you can hang the weight from the end of the beam supported by the wall. m 2ur Additional Materials Reading 3-51 PM O Type here to search fdy 11/14/2017 ^Explanation / Answer
Horizontally, for tension T and normal force on beam Fn, we've got
Fn = Tcos30º = 0.866T
We know that at the threshold, Ff = µ*Fn = 0.440 * 0.866T = 0.38104*T
where Ff is the friction force at the beam
Vertically, we've got Ff + Tsin30º = 0.381*T + 0.5T = 0.881*T = w + 2w = 3w
so T = 3.405*w
Finally, consider the moment about the left end of the beam.
It must be zero, or the beam would rotate.
M = 0 = Tsin30º * L - w*L/2 - 2w*x = 3.405*w * 0.5 * L - w*L/2 - 2w*x
w cancels from all terms:
1.2025*L = 2x
x = 0.6013*L
Since L = 3.75 m,
x = 2.25 m
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