Statics problem(chapter 2 of the book Vector Mechanics for Engineers)

[Moayad]

Homework Statement

Three forces are applied to a bracket. The directions of the two
30-lb forces can vary, but the angle between these forces is always 50. Determine
the range of values of  for which the magnitude of the resultant of
the forces applied to the bracket is less than 120 lb.

http://www.gefoon.net/upload/7/278933efb0.jpg

2. Relevant equation:

Fnet,x=o
Fnet,y=0

The Attempt at a Solution

i tried to get this problem solved by using different ways, but none of these have given me any answers.
@= the angle between 30-lb forces
i tried this one :

Fnet,x= 100 cos(30)-30 cos(@)-30cos(@+50)=0
Fnet,y=100sin(30)-30sin(@)-30sin(@+50)=0

but, this does not seem right for me because i am not going to have range of the angle and i did not use 120 ,and alos the TI-89 did not give me any thing.

however, i used another way,

120> -100 cos(150)+30cos(150)+30cos($)+30cos($+50)

i tried to solve by Ti-89 but also nothing happned.

$= the angle between the x-positive axis to the first 30-lb force.






------------


my problem is i do not know how to deel with (the which the magnitude of the resultant of
the forces applied to the bracket is less than 120 lb.) ...


can anyone helps me with ASAP

 

[tiny-tim]

Welcome to PF!

Hi Moayad! Welcome to PF!

(have a theta: θ )

… i tried to get this problem solved by using different ways, but none of these have given me any answers.
…
Fnet,x= 100 cos(30)-30 cos(@)-30cos(@+50)=0
Fnet,y=100sin(30)-30sin(@)-30sin(@+50)=0

It's Fnet_x = 100 cos(30)+30 cos(@)+30cos(@+50)=0

but, this does not seem right for me because i am not going to have range of the angle and i did not use 120 ,and alos the TI-89 did not give me any thing.

Hint: F² = F_x² + F_y²

(oh … and it might be easier to combine the two 30lbs forces into one force at θ + 25º first )

 

[Moayad]

thanks man ...



can i ask more questions

It's Fnetx = 100 cos(30)+30 cos(@)+30cos(@+50)=0

is not Fnet,x = -100 cos(30)+30 cos(@)+30cos(@+50)=0
what about Fnet,y?

is it goin to be = 100sin(30)+30sin(@)+30sin(@+50)?


Thank you for help


Moayad was here

 

[tiny-tim]

is not Fnet,x = -100 cos(30)+30 cos(@)+30cos(@+50)=0
what about Fnet,y?

is it goin to be = 100sin(30)+30sin(@)+30sin(@+50)?

(what happened to that θ i gave you?)

if x is to the right, it's Fnet,x = -100 cos(30)-30 cos(@)-30cos(@+50)

if x is to the left, it's Fnet,x = 100 cos(30)+30 cos(@)+30cos(@+50)

(where did your "=0" come from? )

Your original Fnet,y was correct.

 

[Moayad]

from the graph


is not the x to the right ?



sorry i did not use what you gave me, i' am going to use from now




thank you again

 

[Moayad]

here again

i did this :

i used the x to the right, so

Fx=-100cos(30)-30cos(25-θ) , ( i combined the two 30lbs forces into one as you told me )
Fy=100sin(30)-30sin(25+θ)

F^2 = Fx^2 + Fy^2

l let F=120 ...(is it right?)

so,

Fx= sqrt((120)^2+Fy^2)

i pulged to to(Fx=-100cos(30)-30cos(25-θ) )

it became :
sqrt((120)^2+Fy^2)=-100cos(30)-30cos(25-θ)

and i pulged (Fy=100sin(30)-30sin(25+θ))

so,

my final equation is :


sqrt((120)^2+(100sin(30)-30sin(25+θ))^2)=-100cos(30)-30cos(25-θ)


then , i put it to the Ti-89 using the solve comand. i solved to the θ

but, it did not gave me a number or range of values

 

[tiny-tim]

from the graph

is not the x to the right ?

Hi Moayad!

i agree x is usually to the right …

but it's your decision …

the maths is only there to help you …

you're in control!

you can make x to the left if you want to, and in this case it's much easier, because it avoids all those minus signs!

 

[tiny-tim]

i did this :

i used the x to the right, so

Fx=-100cos(30)-30cos(25-θ) , ( i combined the two 30lbs forces into one as you told me )
Fy=100sin(30)-30sin(25+θ)

F^2 = Fx^2 + Fy^2
…

Hi Moayad!

hmm …

i] you can't just write 30cos(25-θ) … the magnitude of two 30lb forces combined won't still be 30 (or even 60), will it?

ii] don't be so complicated …

just square Fx and Fy, add them, and use cos²θ + sin²θ = 1

 

[Moayad]

Hi again ...

i] you can't just write 30cos(25-θ) … the magnitude of two 30lb forces combined won't still be 30 (or even 60), will it?

i usually make stupid mistakes ... sorry heh



i added them so what i got is 45.69 lb

just square Fx and Fy, add them, and use cos²θ + sin²θ = 1

i am going to work this out .. and i'll be back



thank you sooooo much

 

[tiny-tim]

i added them so what i got is 45.69 lb

how did you get that?

you didn't use 50º, did you?

 

[Moayad]

i used

30/sin(25)=R/sin(40)

R= the new force

 

[tiny-tim]

i used

30/sin(25)=R/sin(40)

R= the new force

wherever did 40º come from?

resolve in the direction of R: R = 30cos(25º) + 30cos(-25º) = … ?

 

[Moayad]

i got it ... thank you for you help


i used the three equation

1- F=sqrt((Fx)^2+(Fy)^2)

F=120


2-Fx = -100 cos(30)-30 cos(θ)-30cos(θ+50)

3-Fy=100sin(30)-30sin(θ)-30sin(θ+50)


i pluged equation 2,3 to 1
and solve for θ


i got 27.39>θ>223