Why Is Calculating Rope Tension with Angles Confusing?

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Calculating rope tension with angles can be confusing due to the geometry involved. In a scenario with a 20 Newton force hanging from a rope that dips 25 degrees, the tension is derived from the equation 2TCos65 = 20N, where 65 degrees is the angle from the vertical. The discussion highlights a misunderstanding regarding the correct angles and signs in the equations used to solve for tension. The correct approach involves recognizing that the vertical components of tension from both sides of the rope must sum to balance the downward force. Clarifying these points resolves the confusion and confirms that the tension calculated should be approximately 24 Newtons.
druidwoody
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This homework problem was originally posted in another forum so there is no template
Hi,

I have a revision book that has an example in it for the tension on a rope
if a 20 Newton force is hanging on the centre of a rope suspended between two points.

The rope dips 25 Degrees either side. So effectively an upside down pair of right angled triangles is formed.
The Hypoteneuse being 25 Degrees down.

http://d:\Rope-example.png

The Revision textbook Example shows how to find the Tension
2TCos65(Degrees)=20N
and shows T works out to be 24 Newtons.

I can not see where the 2TCos65=20 comes from, at least its not intuitive to me.
So I figured it out with a simultaneous equation. T1 = left T2 = right
Where the forces are worked out, asuming everything is static.

1) T1Cos25 + T2Cos65=0
2) T1Sin65 + T2Sin25=20N

From 1)
3) T1=T2x0.466

From 2)
T1x0.966 + T2x0.422 = 20

Substitute for T1 from 3)
T2x0.466x0.906 + T2x0.422 = 20
=> T2 x 0.844 = 20
Therfore T2 = 23.69N which is close enough to 24 Newtons in the example. I assume this is correct? (Maybe the book is wrong too?)

I was pleased with myself, until I checked by substituting T2 back fot T1 to check T1 was the same. I assume T1 and T2 are equal tensions.
But it is not, which I found very puzzling. I get about 11 Newtons. i.e put 23.69 in place of T2 in 3) above.

Can anyone explain what I have done wrong? Signs perhaps?, unfortunately my math is not the greatest.

Regards,
 

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Your Eqns. 1 and 2 are incorrect. Check the geometry more carefully. All the angles in both these equations should be 25. Also, in Eqn. 1, there should be a minus sign, not a plus sign.
 
druidwoody said:
I can not see where the 2TCos65=20 comes from, at least its not intuitive to me.

It should be.

Step by step...
Its a static problem so the vertical forces sum to zero.
I assume you know where the 65 degree angle is.
The tension each side is T.
The vertical component of each T is TCos65.
There are two of them so the total upward force is 2TCos65.
The downward force is 20N.
Sum the forces to zero...
2TCos65 - 20 = 0
Rearrange to give
2TCos65 = 20
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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