Question

The rigid beam, AB, has a weight of 1.05 kN hanging vertically at point B. The beam weighs 1.20 kN and is supported by a hinge at point A and a cable at point B. Determine the tension in the cable and the reaction forces on the beam at point A.

Answer 1

its just trig, or even simpler, congruent triangles

Answer 2

I am having trouble knowing where and how I should take the moment

Answer 3

I have F\[F _{y}=0=Tsin(\theta)+B _{y}-w _{1}-w _{2}\]

Answer 4

this should allow us to determine the tension in the cable

Answer 5

is your issue with the reaction forces?

Answer 6

and \[F _{x}=0=-Tcos(\theta)+B _{x}\]

Answer 7

sin(theta) is just a ratio of sides, and we know the sides already so theta is immaterial at the moment

Answer 8

so should I take the moment at point A?

Answer 9

yes

Answer 10

which should be the solutions to 1.5 n and 2 n

Answer 11

so \[M _{A}=0=-w _{1}(.75)-w _{2}(2)+2T\]

Answer 12

so T=1.5 kN
\[B _{x}=1.2 kN\]
\[B _{y}=1.35 kN\]

Answer 13

is that correct?

Answer 14

n = 1.05/1.5

such that T = 1.05 sqrt(1.5^2 + 2^2)/1.5 = 1.75 is what im getting

Answer 15

chk that im not violating any dimension rules of course

Answer 16

the meters cancel and we are left with kN

Answer 17

the forces at A are simply 1.5n and 2n given that n=1.05/1.5 seems fair to me

Answer 18

I don't understand where you are getting this n from?

Answer 19

you do understand that sin(theta) = y/r = 1.5/sqrt(1.5^2 + 2^2)

Answer 20

yes

Answer 21

we are given this as the dimensions for our right triangle, the same right triangle will be sclaed such that 1.5 times some value is equal to 1.05 as given

Answer 22

but that 1.05 is a force, not a length?

Answer 23

doesnt matter, the angles are the same and therefore the similar triangles all have the same ratios

Answer 24

but are you forgetting about the weight of the beam?

Answer 25

let n = 1.05 kN/ 1.5 m

therefore: 1.5 m * 1.05 kN/1.5 m = 1.05 kN

Answer 26

you may be right .... does the information given tell us that 1.05 is a result of the beams weight?

Answer 27

no, the weight of 1.05 kN hangs vertically from point B and the beam weighs 1.20. I don't think the 1.05 is part of the beam.

Answer 28

then my idea is prolly flawed since i was assuming the 1.05 was an effect given by the beams weight being in that position to start with

Answer 29

now that i have doubts raised, thanks lol, im not sure that i can proceed with a solution process. math is more natural to me than physics per se

Answer 30

ratio the beams weight, and then add in the vertical?

Answer 31

so you don't think the way I solved it will work?

Answer 32

im not familiar with your formulas so i cant say for sure

Answer 33

well it's a statics problem so all the forces in the x-direction have to sum to 0, same as the forces in the y-direction

Answer 34

right

Answer 35

does that look like a reasonable stick model?

Answer 36

what force is the horizontal arrow on the far right?

Answer 37

well, the beam wants to fall down, the only direction it can want to move at b is down and right

Answer 38

its the forces that are keeping tension on the cable

Answer 39

from my own faulty perspective of course :)