Can Twine Support a 15-N Picture Without Breaking?

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Homework Help Overview

The discussion revolves around the problem of determining whether a 15-N picture can be hung using fine twine that breaks under more than 12 N of tension. Participants are exploring the implications of tension in different configurations as illustrated in the problem statement.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to calculate the tensions in the twine using equilibrium equations and are questioning how to assess whether the tensions exceed the breaking limit of the twine. There is also discussion about whether to consider total tension or only vertical components when evaluating the ceiling's capacity.

Discussion Status

Some participants have provided insights regarding the maximum tension the twine can withstand and the implications for the ceiling's load capacity. There is ongoing exploration of the relationship between total tension and vertical components, with no explicit consensus reached on the best approach.

Contextual Notes

Participants are navigating the constraints of the problem, particularly the maximum tension limit of the twine and the conditions under which the ceiling's support is relevant. There is uncertainty about how to apply these constraints in different scenarios presented in the problem.

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.