Question

A 420-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 130-N rod as indicated in the figure below. The left end of the rod is supported by a hinge and the right end is supported by a thin cable making a 30.0

Answer 1

Given that the weight of the box is W = 420N
        Length of rod is L= 6.00 m
        Weight of rod is Wr =130 N
        Angle with vertical is? = 30^o
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The wieght of the box acts at its center that is at2.00m from each end ( At point C)

Weight of rod acts at center of the rod that is at 3.00mfrom each end ( at poitnt B)

    Apply Torque about the point A

        Fy*0 - Wr*AB- W*AC + T sin60^o*AD = 0

     

                     Wr*AB + W*AC  = Tsin60^o*AD  

          Wr*(3.00m) + W*4.00m  = T sin60^o*6.00m  

                                            T = [ Wr*(3.00m) + W*4.00m  ] / sin60^o*6.00m              

           then we get T = 398.37 N

b)

Apply Newtons law in verticle direction

                   Fy - Wr - W + T sin60^o = 0

                                                   Fy = Wr + W - T sin60^o

                                                          =205 N

This is the vertical force at the hing

a)

Apply Newtons law in horizontal direction

                        Fx = T cos60^o  

                             = 198.185 N