Can Twine Support a 15-N Picture Without Breaking?

AI Thread Summary
The discussion centers on whether a 15-N picture can be hung using twine that breaks at over 12 N tension. Calculations show that T1 reaches 13.27 N and T2 is 9.82 N, indicating that T1 exceeds the twine's limit, which would cause it to break. Participants clarify that the critical factor is the tension in the twine itself, not the ceiling's load capacity. For part B, the tensions will differ, and the same values from part A should not be assumed applicable. The conclusion emphasizes that any tension exceeding 12 N will result in failure of the twine.
SakuRERE
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Homework Statement


you want to hang a 15-N picture as in part (a) using some very fine twine that will break with more than 12 N of tension can you do this? what if you have it as illustrated in part (b) of the figure?
upload_2018-10-6_10-10-29.png

Homework Equations


Σf=ma

The Attempt at a Solution


[/B]
okay i started with finding the tensions in part (a). using:
Σfx= 0 and so
T2 cos30=T1 cos50
T2=0.74 T1 ---------(1)
Σfy=0
T1sin50 + T2sin30= W
0.76T1+0.5T2 = 15-------(2)

substituting (1) in (2)

0.76T1 +0.5 (0.74 T1)=15
0.76 T1 + 0.37 T1 =15
1.13 T1= 15

T1= 13.27N
and so T2= 9.82 N

Now what i am not sure about, is first how to know if the ceiling is fine with these tensions or not? I mean should I only find the vertical forces of the tensions here (T1sin50 & T2 sin 30) and see if they are larger than 12 ( obviously they are ) or what?

and regarding the part B. I am also hesitated about using the same Tension values computed from the first part (a) in b also!

thanks in advance
 

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SakuRERE said:

Homework Statement


you want to hang a 15-N picture as in part (a) using some very fine twine that will break with more than 12 N of tension can you do this? what if you have it as illustrated in part (b) of the figure?
View attachment 231773

Homework Equations


Σf=ma

The Attempt at a Solution


[/B]
okay i started with finding the tensions in part (a). using:
Σfx= 0 and so
T2 cos30=T1 cos50
T2=0.74 T1 ---------(1)
Σfy=0
T1sin50 + T2sin30= W
0.76T1+0.5T2 = 15-------(2)

substituting (1) in (2)

0.76T1 +0.5 (0.74 T1)=15
0.76 T1 + 0.37 T1 =15
1.13 T1= 15

T1= 13.27N
and so T2= 9.82 N

Now what i am not sure about, is first how to know if the ceiling is fine with these tensions or not? I mean should I only find the vertical forces of the tensions here (T1sin50 & T2 sin 30) and see if they are larger than 12 ( obviously they are not) or what?

and regarding the part B. I am also hesitated about using the same Tension values computed from the first part (a) in b also!

thanks in advance
The string can withstand 12 N tension at maximum. You got more than 13 N tension for one of the strings. What would happen to it?
Case b is entirely different, the tensions are different, too.
 
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ehild said:
ou got more than 13 N tension for one of the strings
okay, i got what you said, but the point that i want to ask still somehow not clear.
for example if i got for T2=5 N and for T1= 8N, and the ceiling is fine with 12 maximum.
do i say
T2 + T1= 5+8= 13 so no! it's more than the limit
or do i take only vertical components of the tensions like saying:
T2 cos 30 + T1 cos 50 =?
5 cos30 + 8 cos50 = 9.47 so the ceiling is fine with it?
Thanks
 
SakuRERE said:
okay, i got what you said, but the point that i want to ask still somehow not clear.
for example if i got for T2=5 N and for T1= 8N, and the ceiling is fine with 12 maximum.
do i say
T2 + T1= 5+8= 13 so no! it's more than the limit
or do i take only vertical components of the tensions like saying:
T2 cos 30 + T1 cos 50 =?
5 cos30 + 8 cos50 = 9.47 so the ceiling is fine with it?
Thanks
It is not the ceiling that is critical. The problem days "using some very fine twine that will break with more than 12 N of tension can you do this? " The twine will brake if overloaded! The tension in any of them can not exceed 12 N.
If the problem said that the ceiling can not withstand more than 12 N load, you might use the sum of the vertical components of the tensions.
 
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