Question

My answer is wrong. Can anyone check my solution? I got approx, 490 N but the correct answer is 550N. Why is that? I'll attached both the question and my solution.

Answer 1

@chris00 physics?

Answer 2

is the answer highlighted in yellow?

Answer 3

yeah

Answer 4

okay lets do this

Answer 5

lets get forces in terms of their x and y components

Answer 6

F1 first...

F1x= 850(4/5)=680
F1y= 850(3/5)=510

right?

Answer 7

yep or we can write it as\[F _{1}=\left[ 680i-510j \right] {\space}kN\]

Answer 8

now lets do F(2)

Answer 9

oh okay..

next f2:
F2x= 625sin(30)=312.5N
F2y=625cos(30)=541.3N

oh i think this is the part i got wrong

Answer 10

\[F _{2}=\left[ (-625\sin30)i-(624\cos30)j \right] kN\]

Answer 11

remember unit circle

Answer 12

so in the third quadrant, only tan is positive, hence since and cos are negative

Answer 13

\[F _{2}=\left[ -312.5i-541.3j \right]{\space} \]

Answer 14

now do F3

Answer 15

okay..

F3:
F3x=750 sin(45)=530.33 N (-x direction)
F3y=750cos(45)=530.33 N (+y direction)

so if i'll put it like your way it would be like (-530.33i, 530.33j) ?

Answer 16

yep!

Answer 17

so lets write out full equations out

Answer 18

\[F _{1}=\left[ 880i-510j \right]{\space} kN\]
\[F _{2}=\left[ -312.5i-541.3j \right]{\space} kN\]
\[F _{3}=\left[ -530.3i+530.3j \right]{\space} kN\]

Answer 19

Now, \[F _{R}=\sum_{}^{}F _{i}+\sum_{}^{}F _{j}\]

Answer 20

add all the x and all the y components

Fnet= (-162.8i, 520.6j)

Answer 21

then get the magnitude of Fnet: sqrt( (162.8)^2 + (520.6)^2)=545.5 N or 550 N

:D thanks!

Answer 22

\[F _{R}=(680-312.5-530.3)i+(-510-541.3+530.3)j\]
\[F _{R}=-162.8i+520.6j\]

Answer 23

yep, spot on!

Answer 24

haha why did you choose me to do this question? haha

Answer 25

oops sorry late response.

I choose you because you're the only one online that I fanned
and you're nice so yeah xD thanks again!