Question

A rectangular sign over a store has a weight of 290 pounds, with the center of gravity in the center of the rectangle. The support against the wall at point C may be treated as a ball and socket joint. At corner D support is provided in the y direction only. Calculate the tensions T1 and T2 in the supporting wires, the total for supported at C1, and the lateral force R supported at D. The force R is positive if it points in the +y-direction, and negative if it points in the -y-direction.

5 7.5 C 3.0 TOYS B.O 12.0

Answers for

T1

T2

C=228

R=

i have posted once already. the only one right was C.

Answer 1

The coordinates of points are: A(0,-4.5, 7.5) B(0,4.5, 7.5) C(0,0,-3) D(0,0,0) Assume cable from B is connected at E on board and cable from A is connected to F on board E (7.5,0,0) F(12,0,0) Center of gravity of board, G(6,0,-3) Write the forces in the vector form: W =-290k Fe = Foxi +FcvJ+FCK 7p = Faj FEB = Tz (Unit vector in direction of EB) (0-7.5)ī +(4.5-0))+(7.5-0) (10-7.5) +(4.5–0)*+(7.5–0), = T2(-0.65097 +0.39063 +0.65092) FEA=T(Unit vector in direction of FĀ) (0-12)ī +(-4.5-0); +(7:5–0) Wo-12)2+(-4.5-0)*+(7.5-0)") = T (-0.80817 – 0.3031 +0.5051K)

Apply the force balance equation. F=7 FrA +FEB +Fp+Fc+W = 7 (FC -0.6509T2 -0.808177 )i + (Fo +F4 +0.3906T2 -0.3031,); +(Fc - 290 +0.6509T2 +0.50517,)k = 0 Equate the individual components to zero. FG -0.65091, -0.80817, = 0 Fo+Fq+0.3906T2 -0.303T, = 0 Fc - 290 +0.65097, +0.50517= 0 Equate the moments about the point D to zero. M = Text +7cxFc+ię x FgB+TE X FA= 60 -3 + 0 0 -3 +12 7. 5 0 +T 12 0 0 = 7 LO O -290] [Fca Fey Fc] [-0.6509 0.3906 0.6509 | -0.8081 -0.303 0.5051 (3F,)7-(-6* 290 + 3F+7.5*0.65097, +12*0.50517, )j +(7.5*0.3906T, – 12*0.3037, )= Equate the individual components to zero. Fo=0 -6* 290 + 3FG + 7.5*0.65097, +12*0.50517; = 0 7.5*0.3906T2 -12*0.30317 = 0 Solve the 6 equations for the 6 unknowns. FG = 165.7;FG = 0; Fc. = 155.4 Fa=-18.64;7; = 102.5;T2 =127.3 Fo=F&+F+F2 = 227.1