A rectangular placard weighing F, has a width of 2L. The placard is suspended from a light, horizontal rod hinged at the wall and supported by a cord.Lulu and Lisa'sCafe(a) Determine the tension in the cord. (Use any variable or symbol stated above as necessary.)T =(b) Determine the components of the reaction force (R) exerted by the wall on the rod, in terms of Fa, d, L, and e. (Assume the positive horizontal direction is to the right and the positive vertical direction is upward. Indicate the direction withthe signs of your answers.)horizontalR, =verticalRy =

Given: Weight of placard F g Width of placard 2L The positive horizontal direction is to the right and the positive vertical direction is upward. To find: (a) Tension in the cord (b) Components of the reaction force exerted by the wall on the rod. Answer(a): Formula used: l=(d+2L)sinθ where 'l' is the arm length 'd' distance from placard's left support to edge '2L' placard length 'θ' support cord angle with the horizontal Calculation: Net torque around 'O' is calculated as:-(Fg2)(d)-(Fg2)(d+2L)+T(l)=0As l=(d+2L)sinθwe get:-(Fg2)(d)-(Fg2)(d+2L)+T((d+2L)sinθ)=0T((d+2L)sinθ)=Fgd+FgLT=Fg(d+L)sinθ(d+2L)Answer(a) is Tension in the cord is T=Fg(d+L)sinθ(d+2L) Answer(b): Calculation: Net force in the horizontal direction:Rx-Tcosθ=0using Fg(d+L)sinθ(d+2L) for T Rx-Fg(d+L)sinθ(d+2L) cosθ=0Rx=Fg(d+L)cotθ(d+2L)Net force in the vertical direction:Ry+Tsinθ-Fg=0using value of TRy+Fg(d+L)sinθ(d+2L)sinθ-Fg=0Ry=Fg(d+2L)-Fg(d+L)d+2LRy=FgLd+2LAnswer(b) is Horizontal Rx=Fg(d+L)cotθ(d+2L) and Vertical Ry=FgLd+2L

Answered: A rectangular placard weighing F, has a…