Chapter 12, Problem 030 In the figure, a 56.7 kg uniform square sign, of edge le

Question

Chapter 12, Problem 030 In the figure, a 56.7 kg uniform square sign, of edge length L 1.77 m, is hung from a horizontal rod of length dh 3.42 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance di 4.13 m above the point where the rod is hinged to the wall. (a) what is the tension in the cable? (b) hat is the horizontal component of the force on the rod from the wall? Take the positive direction to be to the right. (c) What is the vertical component of this force? Take the positive direction to be upward. Cable llinge Rod H. Per (a) Number (b) Number (c) Number sasow HINT Go TurORLAL I A2000-2017 2ohn Wiley sons, Inc. All Rights Reserved. A Division of lohn Wiley & Sons Inc. 4/23/2017

Explanation / Answer

mass of the uniform sign m = 56.7 kg

length of the uniform sign L = 1.77 m

length of the rod , dh = 3.42 m

the distance between wall and rod , dv = 4.13 m

from figure ,

    tan = dv/dh

            = (4.13m) / (3.42m)

angle = tan-1(1.207)

           = 50.37 degree

from equilibrium condition , the net torque is

     = (T sin)(dh) - (mg)(dh - L/2) = 0

             (T sin)(dh) - (mg)(dh - L/2) = 0

therefore ,

A. tension in the cable is

      T = [(mg)(dh - L/2)] / (dh) sin

         = [(56.71 kg)(9.8 m/s2)(3.42 m - 0.885 m)/3.42×sin (50.37)

= 534.867 N

B. horizontal component of force

   Fx = T cos(50.37)

        = 341.152 N

ddirection : toward right

C. vertical component of force

      Fy = mg - T sin(50.37)

           = 143.814 N

direction : upward direction

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