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Tension in a rope at an angle with a hanging mass

  • Thread starter Wolfmannm
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  • #1
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Homework Statement


A rope is connected between two fixed points that are separated horizontally from one another by 10 m. The point on the left at a height of 5 m [coordinates (0,5m)], and point on the right is at a height of 10 m [coordinates
(10m,10m)]. A 100 lbs weight hangs from the rope, attached at the location (5m,6m), separating the rope into two segments. What is the tension in each segment? Neglect the weight of the rope. See attached image

2. Homework Equations
T1 = mg sin ϴ1
T2 = mg sin ϴ2



The Attempt at a Solution


I fist converted the 100 lb weight to 444.8N. I was thinking that the problem would be similar to one with a horizontal rope with the ends at equal heights, and solved for both ϴ1 and ϴ2 then used these values to calculate the tension in each segment. I got T1 = 87.2 N and T2 = 278.1 N.
Does this seem correct?
 

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Answers and Replies

  • #2
collinsmark
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Hello Wolfmannm,

Welcome to PF! :)

Homework Statement


A rope is connected between two fixed points that are separated horizontally from one another by 10 m. The point on the left at a height of 5 m [coordinates (0,5m)], and point on the right is at a height of 10 m [coordinates
(10m,10m)]. A 100 lbs weight hangs from the rope, attached at the location (5m,6m), separating the rope into two segments. What is the tension in each segment? Neglect the weight of the rope. See attached image

2. Homework Equations
T1 = mg sin ϴ1
T2 = mg sin ϴ2



The Attempt at a Solution


I fist converted the 100 lb weight to 444.8N. I was thinking that the problem would be similar to one with a horizontal rope with the ends at equal heights, and solved for both ϴ1 and ϴ2 then used these values to calculate the tension in each segment. I got T1 = 87.2 N and T2 = 278.1 N.
Does this seem correct?
Those answers (87.2 N and 278.1 N) are not correct. There must be a mistake in there somewhere.

Post your steps on how you obtained your answers, and maybe we can help pinpoint what went wrong.

[Edit: And also, the equations that you listed in your relevant equations section are incorrect for this problem.]
 
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  • #3
CWatters
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Can you show your working? T2 sounds a too low, it must be > 444.8N.

I haven't worked right through to the answer (it's 11pm here) but I have four equations and four unknowns. One of my equations sums the vertical components to zero....

T2Sinϴ2 - 444.8 - T1Sinϴ1 = 0
so
T2 = (444.8 + T1Sinϴ1) / Sinϴ2
or
T2 = 444.8/Sinϴ2 + T1Sinϴ1/Sinϴ2

Sinϴ2 is < 1 so that implies T2 > 444.8N
 
  • #4
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I calculated ϴ1 to be approx -11.3° and ϴ2 to be approx 38.7°. Do these seem reasonable?

I see where those equations are incorrect. I didn't account for forces acting in more than on direction. Is the last equation
T2Cosϴ2 - 444.8 - T1Cosϴ1 = 0 ?
 
  • #5
collinsmark
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I calculated ϴ1 to be approx -11.3° and ϴ2 to be approx 38.7°. Do these seem reasonable?
Those numbers seem reasonable, yes, with the possible exception of the negative sign on the "-11.3o". That depends on how you interpret the angle. Given your diagram, both angles should be positive numbers, given the way you have defined them in the diagram, I would think.

I see where those equations are incorrect. I didn't account for forces acting in more than on direction. Is the last equation
T2Cosϴ2 - 444.8 - T1Cosϴ1 = 0 ?
Something is not quite right here. The tension times the cosine of the angle is the horizontal component of the particular tension, given how you have defined the angles in your diagram. The mass' weight acts in the vertical direction, not the horizontal.

Eventually you will need two simultaneous equations to combine. One for the sum of vertical forces, and the other for the horizontal forces.
 
  • #6
collinsmark
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By the way, if you're ever in doubt about which trigonometric function to use, recall,

[tex] \sin \theta = \mathrm{\frac{opposite}{hypotenuse}} [/tex]
[tex] \cos \theta = \mathrm{\frac{adjacent}{hypotenuse}} [/tex]
[tex] \tan \theta = \mathrm{\frac{opposite}{adjacent}} [/tex]

I'm sure that you probably already know that. But my advice here is to commit those to memory in such a way that they become second nature.
 
  • #7
CWatters
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I calculated ϴ1 to be approx -11.3° and ϴ2 to be approx 38.7°. Do these seem reasonable?
They are reasonable but don't quite satisfy two of my equations. If I have understood correctly the "attached at the location (5m,6m)" means that horizontally the dimensions give you...

5Cosϴ1 + 6Cosϴ2 = 10

If I substitute your values for the angles I get = 9.6

I see where those equations are incorrect. I didn't account for forces acting in more than on direction. Is the last equation
T2Cosϴ2 - 444.8 - T1Cosϴ1 = 0 ?
As colinsmark said.. That's not correct. Are you trying to sum the horizontal or vertical components? It would help if you show us all 4(?) of your initial simultaneous equations with a one line explanation of each.
 

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