Question

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5) 3. The adjacent diagram shows a design for supporting a traffic light (100 kg). It hangs form the end of a 7.5 meter long pole (50 kg). Assume ZACD = 90°. Find the tension in the cable CD and the magnitude and direction of the force exerted at point A. (8 pts) B 37 Hint: CD = m, AD = m 3.80 m A

Answer 1

First things are first: The tension from the cable is not acting at the tip of the pole (as you see that the string is not linarly connected with the tip ), but only partially up the pole and before the hanging stop light. The distance that the cable connects to the hanging pole is the hypotenuese of a triangle which has a side of 3.8 m and opposite angle of 37 degrees, so the length of this hypotenuse (AD) is -

Isin (37) = 3.80m

3.80 sin(37)

l = 6.3 m

now,

we start by considering Newton's and law angular rotation for 37 ting e CAD)= 6.3m. , Mg LF mass of traffic light (m) = 100kg. mass of poe CM) = 50 kg. length of pole = 95 m. Now, lets take our rotation through the base A. 2 : T sỉ (31) 6 3%- (100 và 8 LOS 53 (1.5 km ) - (50 kg yg cos (539( 15.) -... (A) we are taking the force of the weight of the pole oching from its centre of mass (middle since its uniform). we can solve the above equation as - T siu (370) 6:3 m = (100 x 0.601 X 7.5x9) - 50 89.8 X0.60) X3-75) T siu ( 37°) 6.3 m = 4417.35 – 1104.3 T: 3313.01 3.791 T: 873.9 N.

Next, we som the forces in x and y directions and since the system is static, these sums are zero. EF= F,-1=0 so, F, =T F = 873.9N. e Fy = Fa-ng-mg F = (M+m) g = (100+50)g = 15089.8 N = 1450 N. so, The tension "T" on the cable is [873.g. Ans. and Force exerted at A is in æ aud y direction as shown in diagram where - F = [873.90 Fg = 11470 N An.
Hope, it helped.