[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

View post   
File: 44 KB, 450x234, Figure_04_05_06.jpg [View same] [iqdb] [saucenao] [google]
8508935 No.8508935[DELETED]  [Reply] [Original]

A 5 meters long rope is tied to two trees that are 4meters apart. A mass of 70kg is placed on middle of the rope.
a)Calculate tension of the rope.
b)If the tension was 1352N, what would be the shortest length of the rope that could withstand it.

>Are the right answers?:
>a) T=2.5N ( prob wrong by far)
>b) l(min) = 0.8m

>> No.8508940

>>8508935
Are you fucking retarded?

>> No.8508947

>>8508935
Calculate the Lagrangian and let me know what you get so I can help

>> No.8508965

>>8508940
>No, just not good at math
>im perfectly aware how theory works, but dont know how to calculate it
>>8508947
>Not sure what Lagrangian is.

>> No.8508979

>>8508965
As the angle is unknown but we have simmilar triangle i tried with cos(alfa)=Txl/TL(where Txl is x component of TL)

>> No.8508986

Indeed wrong by far. Use the law of sines.

a) T= 3939N

Does the angle stay the same in b)? Im too lazy to think

>> No.8508990

>>8508986
Yes it stays the same, Thank you for advice.

>> No.8508999

>>8508990
Well then your b) is pretty much correct. I got 0,85m.

>> No.8509012

>>8508999
Might be wrong way to do it.
I did it so:
Tcos(alfa)=TxL
and as TxL/T = d/2 / l/2 i got l/2 = TxL / d/2 * T
>Btw. do you get angle of 36.87° ?

>> No.8509029

>>8509012
Yeah uh I just realized my mistake

x/1352 = 5/3939

x = 1,7m

>> No.8509043

>>8508935
1./sci/ is not for doing your homework
2. gtfo back to middle school

>> No.8509132

Did you find the wave function?

>> No.8509147

>>8508935
SQT?

>> No.8509235

>>8508935

[math]70\mathrm{kg}*9.81\mathrm{ms^{-1}}=686.7\mathrm{N}[/math] <-- Weight on rope.
The tensions on either side will cancel out the weight.

By resolving vertically,
[math]686.7\mathrm{N}=2*\mathrm{T}*\sin{5.0\textdegree}[/math]
[math]\mathrm{T}=\frac{686.7\mathrm{N}}{2*\sin{5.0\textdegree}}=3939\mathrm{N}[/math]

>>>/wsr/ for future reference.

>> No.8509264

>>8509235
Thx for /wsr/ info.
But alfa isnt 5° pic is just example (in the assignment angle is unknown) ill try to calculate from how you did it. Thank you

>> No.8509292

>>8509235
Also as i need angle is the right way to do it:
cos(alfa) is half distance devided by half of rope lenght? cos(alfa)= 2/2.5 and when i do arccos(2/2.5) i get alfa= 36.87°

>> No.8509318

>>8509292
d=4m, l=5m, m=70kg, W=70*9.81=686.7N
cos(a)=4/5, sin(a)=3/5, a=36.87°
(W/2)/T=sin(a)
T=W/2sin(a)=572.25 N
l for T=1352N:
sin(a)=W/2T
cos(a)=4/l=cos(asin(W/2T))
l=4/cos(asin(W/2T))
l=4.1356m
a=14.71°
verify carefully

>> No.8509323

>>8509318
Thank you very much :) i managed to get a) part of it, but was confused about b) part.

>> No.8509329
File: 335 KB, 540x405, killme.png [View same] [iqdb] [saucenao] [google]
8509329

>>8509292
>Me not reading properly for the millionth time

[math]\cos{\theta} = \frac{\frac{4}{2}}{\frac{5}{2}} = \frac{4}{5}[/math]
[math]\theta = \arccos{\frac{4}{5}} = 36.87[/math]

[math]686.7N=2∗T∗\sin{36.87}[/math]
T=\frac{686.7N}{2∗sin{36.87}}=572.3N

>> No.8509340

>>8509318
I just noticed now, the weight cant be same if tension changed. So weight needs to change accordingly

>> No.8509348

>>8509340
I might be wrong.. :(

>> No.8509390

kek i remember physics 1

also why hasnt this thread been deleted it has homework in the title

>> No.8509411

>>8509390
its up 3 and half hours. Dont know why

>> No.8509633

>>8509348
Tension also depends on angle.

>> No.8510290

>>8508935
lol fuck you

>> No.8510498

this is like babby's first physics problem
read the textbook more thoroughly before starting on the assignments
you literally have the entire problem modelled in the sketch

>> No.8510511

>>8509633
Exactly. The shorter the rope is, the flatter the angle will be and the bigger the tension will be for the same vertical force.