• Support PF! Buy your school textbooks, materials and every day products Here!

Force question

  • Thread starter holezch
  • Start date
  • #1
251
0

Homework Statement



A sailor tests the tension in a wire rope holding up a mast by pushing against the rope with his hand at a distance L from the lower end of the rope. When he exerts a 'transverse' push N, the rope suffers a transverse displacement s.

Show that for s << L, the tension in the wire rope is given approximately by the formula:

T = N L/s




Homework Equations





The Attempt at a Solution



I looked up the word transverse and it meant to be perpendicular. So, then we have that the tension vector, and the displacement vector and another side make a right angle triangle. Then, T is the hypotenuse of the triangle displacement^2 + another side^2.. I know that if I can get a proper sine or cosine equation, I can use the opp/hyp identities to come to that solution, but I can't see it.. thanks
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
Since s << L, tanθ~ sinθ ~ θ = s/L.
So F/T = sinθ = s/L
 
  • #3
251
0
hi, thank you very much, but how is sin theta = F/T? the displacement s would form a right angle triangle with L and where the rope used to be before it was displaced. the applied force could be of varying magnitudes so we can't say that it's the same as S?

thanks
 
  • #4
251
0
or maybe when theta is really small, both triangles formed by N and T and L and S are approximately the same? thanks
 
  • #5
rl.bhat
Homework Helper
4,433
5
or maybe when theta is really small, both triangles formed by N and T and L and S are approximately the same? thanks
Yes. In a triangle formed by the forces, sides of the triangles are proportional to the corresponding forces.
 
  • #6
251
0
thanks! that;s what I thought intuitively.. but how can I show this?

thanks again
 
  • #7
rl.bhat
Homework Helper
4,433
5
thanks! that;s what I thought intuitively.. but how can I show this?

thanks again
It is a theorem in the vectors. No need to prove it here.
 
  • #8
251
0
It is a theorem in the vectors. No need to prove it here.
thanks, do you know where I can learn about this theorem? I had no idea about it.. and is it a theorem involving vectors in PHYSICS or vectors in general? it seems to rely on the physics behind this question. (that the force vectors will be proportional to the sides L and S) thanks again
 
  • #9
251
0
could anybody tell me about this please?
thanks
 
  • #10
rl.bhat
Homework Helper
4,433
5
Original position of the rope, transverse force and the tension in the rope firm a right angled triangle. In that
F/T = sinθ.
When s <<< L, sinθ = θ. And θ = s/L.
So F/T = s/L.
 
  • #11
251
0
thanks, but how do you know that F/T = sin theta?? what if the applied force vector is longer than the tension, then it would have to be scaled..
 
  • #12
rl.bhat
Homework Helper
4,433
5
thanks, but how do you know that F/T = sin theta?? what if the applied force vector is longer than the tension, then it would have to be scaled..
When you apply transverse force in the horizontal direction, the vertical rope is shifted sideways having a tension T. Draw the diagram and find sinθ.
 
  • #13
251
0
yes, I understand, but how do we know that these vectors (the applied force and the tension) will make a right angle triangle with the rope position before the force was applied?

For example, the tension vector might be "too long", and only part of this vector will be included in the triangle.. I hope you understand what I'm asking.. thanks
 
  • #14
rl.bhat
Homework Helper
4,433
5
Can you draw a figure as described in the problem?
I presume the rope is straight before applying the transverse force. Even if the tension vector is "too long", the tension is same through out the rope.
 
  • #15
251
0
hi,
72zj1u.jpg

I also drew in what I understand from this problem.. there's the triangle, and the lengths in red should form a right angle triangle.
the broken line is where it is before being pushed. thanks

I was asking about how we knew that N, T and the broken red line part formed a "proper" right angle triangle? what if a right angle triangle is only formed with N, the red broken line part and a part of T (3/4? 1/2?. et c), then sin theta wouldn't be N/T
 
  • #16
rl.bhat
Homework Helper
4,433
5
If N is the force of push, the vector N should be from broken line to full line. The broken line, N and T form a right angled triangle. Angle between the N and the broken line is 90 degrees.
 
  • #17
251
0
hmm, that's what I thought when I first saw it.. but how do I know the length of N is s and the length of T is L? Isn't it possible that you would need more force to move displacement s than a magnitude of s? same with tension..

thanks
 
  • #18
rl.bhat
Homework Helper
4,433
5
Since two ends of the ropes are fixed, the change in the length of the rope is very minimum. When you push more, θ increases and hence s increases. T also increases, but L does not change. There must be a braking point for the rope. So you have restrict your N.
 

Related Threads for: Force question

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
878
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
817
  • Last Post
Replies
1
Views
923
  • Last Post
Replies
3
Views
839
Top