Determine the force in cables AB and AC necessary to support...

Determine the force in cables AB and AC necessary to support the 11.5-kg traffic light in (Figure 1). Determine $F_{AB}$ Express your answer to three significant figures and include the appropriate units. Figure B C 12° 1 of 1 > $F_{AB}$ = 225 N Submit Previous Answers Request Answer Incorrect; Try Again; 6 attempts remaining Part B Determine $F_{AC}$ Express your answer to three significant figures and include the appropriate units. $F_{AC}$ = 233 N Submit Previous Answers Correct

Transcript

00:03 So in this question we have been given a situation on the right.

00:08 We have a traffic light which is hanging by a combination of two strings, a .b and b .c.

00:17 Which are given some angles with the horizontal.

00:20 And we have been given that the mass of the traffic light is 35 kg.

00:25 And we are asked to find the tension in the string a .b.

00:29 And b .c.

00:32 So this is a body in or this is a system in in equilibrium it's not moving about so all the forces are in equilibrium with each other so let's draw the free body diagram for the knot b so what that gives us is that this is b and there's this tension let's call it t2 there's this tension it's called t1 and there is this weight w of the traffic light and so we also see from here that if this is also the horizontal so this angle will also be 37 degrees because they're alternate angles this angle and this angle are alternate angles and this will be 53 which is the same case so this angle and this angle are alternate angles so let's write these angles down here so this would be 37 degrees and this would be 53 degrees and also let's give a coordinate system let's say this is positive y is towards the top side and positive x is towards the right side so now to find t1 and t2 we will and also it's mark t1 and t2 here let's say this is t1 in this ab and t2 is in bc so now since this system is in equilibrium all we need to do is we need to sum up the forces in the x direction and the y direction and equate them to zero and from that we find out t1 and t2 so let's do that so first let's say that the summation of forces in the x direction in the direction will be equal to so t2 will have a cost component here cost 37 component along the x direction so that we will call and it's in the positive direction so we call it t2 cost 37 degrees and the t1 component here has a negative value because it's in the minus x direction as we assume so this will be minus t1 cost 53 degrees and this should be equal to zero because the body is under equilibrium and then we solve for this we get 0 .8 t2 is equal to 0 .6 t1 or we can say that t2 is equal to 0 .6 by 0 .8 t1 and this let's say the equation 1.

03:22 Similarly we solve for the summation of forces in the y direction and those will be the two components of t1 and t2, two components and w in the negative direction...

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