Tensions of Strings (HW Check)

  • Thread starter Masrat_A
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  • #1
Masrat_A

Homework Statement


In Figure 5, the two strings attached to the mass are fixed to the ceiling and the wall, respectively. Determine the tensions in the strings.

Figure: http://i.imgur.com/jbTo1Nnh.jpg

Homework Equations


Please look below.

The Attempt at a Solution


Ceiling ##= 37^o##
Wall ##= 137^o##
Object ##= 5 kg##

##F_t = F_g##
##F_t = Mg##
##F_t = 5 * 10##
##F_t = 50N##

Ceiling
##F_1 = 50cos37^o##
##F_1 = 40N##

Wall
##F_1 = 50sin137^o##
##F_1 = 34N##

Does this look reasonable?
 

Answers and Replies

  • #2
andrewkirk
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There are a number of problems:
  • The diagram says the angle between the wall string and the horizontal is 137 degrees, which is obtuse, yet it is an acute angle.
  • The solution uses the symbol ##F_1##, which is undefined.
In the free body diagram, using ##T_c## and ##T_w## for the tension in the ceiling and wall strings respectively, and ##T_{c,x},T_{w,x}## for the horizontal components and ##T_{c,y},T_{w,y}## for the vertical components, you need to write out formulas for each of those four components in terms of ##T_c,T_w## and the two angles.
Then write two equations, one to require the total of all vertical (##y##) components to be zero, and the other for the horizontal (##x##) components.

You will have two unknowns: ##T_c,T_w## and two equations, so you can solve them.
 
  • #3
Masrat_A
The diagram says the angle between the wall string and the horizontal is 137 degrees, which is obtuse, yet it is an acute angle.

I don't have much trigonometry background, unfortunately; could you please expand on this a little more? Does this mean I cannot be using sine, and if not, what would you recommend me to do?

The solution uses the symbol ##F_1##, which is undefined.

In the free body diagram, using ##T_c## and ##T_w## for the tension in the ceiling and wall strings respectively, and ##T_{c,x},T_{w,x}## for the horizontal components and ##T_{c,y},T_{w,y}## for the vertical components, you need to write out formulas for each of those four components in terms of ##T_c,T_w## and the two angles.
Then write two equations, one to require the total of all vertical (##y##) components to be zero, and the other for the horizontal (##x##) components.

You will have two unknowns: ##T_c,T_w## and two equations, so you can solve them.

Could you please explain to me what the formulas for x and y would be? Afterward, in what way shall I set the vertical and horizontal components to zero?
 
  • #4
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I don't have much trigonometry background, unfortunately; could you please expand on this a little more? Does this mean I cannot be using sine, and if not, what would you recommend me to do?
From the figure, it looks to me like the angle is 37 degrees, not 137 degrees. I think what is being interpreted as a "1" is actually the angle indicator - the same as is the case in Figure 1.
 
  • #5
Masrat_A
From the figure, it looks to me like the angle is 37 degrees, not 137 degrees. I think what is being interpreted as a "1" is actually the angle indicator - the same as is the case in Figure 1.

Would this then allow me to use cosine on the angle?
 
  • #6
Nidum
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upload_2017-9-11_21-1-2.png


Yes . . Both angles seem to be the same at 37 degrees .
 
  • #7
Masrat_A
In the free body diagram, using ##T_c## and ##T_w## for the tension in the ceiling and wall strings respectively, and ##T_{c,x},T_{w,x}## for the horizontal components and ##T_{c,y},T_{w,y}## for the vertical components, you need to write out formulas for each of those four components in terms of ##T_c,T_w## and the two angles.

Would these be the vertical and horizontal components for ##T_c## and ##T_w##?

##T_{c,x} = 50cos{37} = 39.93##
##T_{c,y} = 50cos{37} = 30.09##

##T_{w,x} = 50cos{37} = 39.93##
##T_{w,y} = 50cos{37} = 30.09##

Then write two equations, one to require the total of all vertical (##y##) components to be zero, and the other for the horizontal (##x##) components. You will have two unknowns: ##T_c,T_w## and two equations, so you can solve them.

I'm still a little confused on this however. Could you please expand on it a little more?
 
  • #8
haruspex
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Would these be the vertical and horizontal components for ##T_c## and ##T_w##?

##T_{c,x} = 50cos{37} = 39.93##
##T_{c,y} = 50cos{37} = 30.09##

##T_{w,x} = 50cos{37} = 39.93##
##T_{w,y} = 50cos{37} = 30.09##
?
No. By what law of physics do you get those equations?

If the overall tension in the ceiling attachment is Tc, what is its horizontal component?
What horizontal forces act on the mass? For equilibrium, what equation can you write?
 
  • #9
Masrat_A
I'm sorry; I misinterpreted what was written on my textbook. Upon further inspection, here is what I have ended up with so far. Do any of these seem right?

For the string attached to the wall, our degree is 37; there is a 90 degree angle formed by the dotted lines in our figure above, and 37 subtracted from 90 gives us 53.

##T_1cos37^o = T_2cos53^o##
##T_1sin37^o+T_2sin53^o = Fg##

##Fg = Mg##
##Fg = 5 * 10##
##Fg = 50##

##T_2 = T_1cos37^o/cos53^o##
##T_2 = 1.327T_1##

##T_1sin37^o+(1.327T_1)sin53^o = 50##
##T_1(sin37^o+1.327*sin53^o) = 50##
##1.662T_1 = 50##
##T_1 = 30.084##

##(30.084)cos37^o = T_2cos53^o##
##24.026 = T_2(0.602)##
##T_2 = 39.9103##

##T_1 = 30.084N##
##T_2 = 39.9103N##
 
  • #10
haruspex
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I'm sorry; I misinterpreted what was written on my textbook. Upon further inspection, here is what I have ended up with so far. Do any of these seem right?

For the string attached to the wall, our degree is 37; there is a 90 degree angle formed by the dotted lines in our figure above, and 37 subtracted from 90 gives us 53.

##T_1cos37^o = T_2cos53^o##
##T_1sin37^o+T_2sin53^o = Fg##

##Fg = Mg##
##Fg = 5 * 10##
##Fg = 50##

##T_2 = T_1cos37^o/cos53^o##
##T_2 = 1.327T_1##

##T_1sin37^o+(1.327T_1)sin53^o = 50##
##T_1(sin37^o+1.327*sin53^o) = 50##
##1.662T_1 = 50##
##T_1 = 30.084##

##(30.084)cos37^o = T_2cos53^o##
##24.026 = T_2(0.602)##
##T_2 = 39.9103##

##T_1 = 30.084N##
##T_2 = 39.9103N##
Yes, except you haven't said which is T1 and which T2.
 
  • #11
Masrat_A
Oh, I'm sorry. T1 is the ceiling, and T2 is our wall.
 
  • #12
haruspex
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Oh, I'm sorry. T1 is the ceiling, and T2 is our wall.
Which would you expect to have the greater tension?
 
  • #13
Masrat_A
I would imagine the ceiling to have greater tension.
 
  • #14
haruspex
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I would imagine the ceiling to have greater tension.
So what do you think you might have done wrong?
 
  • #15
Masrat_A
So what do you think you might have done wrong?

Perhaps I've gotten the degrees the other way?
 
  • #16
haruspex
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Perhaps I've gotten the degrees the other way?
So let's check your first equation in post #9, the horizontal balance of forces. What is the horizontal component of T1?
 
  • #17
Masrat_A
Should it have been ##T_1 = vcosϴ## rather than ##T_1 = cosϴ##? I apologize; I'm having a very tough time pinpoint precisely what went wrong.
 
  • #18
haruspex
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Should it have been ##T_1 = vcosϴ## rather than ##T_1 = cosϴ##? I apologize; I'm having a very tough time pinpoint precisely what went wrong.
Neither of those suggestions make sense. A component of T1 would be T1 multiplied by the sine or cosine of some angle. Previously you multiplied it by cos(37°). Was that right?
 
  • #19
Masrat_A
No, that was not right. I feel like I'm understanding what's happening here. In our figure, ##T_1##, which is attached to the wall, is stated to be ##37^o##. Between both strings, there is a ##180^o## angle; therefore, ##T_1##, our ceiling, would be ##180 - 37 - 53 - 37 = 53^o##.

##T_1cos53^o = T_2cos37^o##
##T_1sin53^o+T_2sin37^o = Fg##

##Fg = Mg##
##Fg = 5 * 10##
##Fg = 50##

##T_2 = T_1cos53^o/cos37^o##
##T_2 = 0.754T_1##

##T_1sin53^o+(0.754T_1)sin37^o = 50##
##T_1(sin53^o+0.754*sin37^o) = 50##
##1.252T_1 = 50##
##T_1 = 39.936##

##(39.936)cos53^o = T_2cos37^o##
##24.034 = T_2(0.799)##
##T_2 = 30.080##

##T_1 = 39.936N##
##T_2 = 30.080N##
 
  • #20
haruspex
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That's it!
 

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