Solving Tension on Strings in Diagram

  • Thread starter Thread starter max0005
  • Start date Start date
  • Tags Tags
    Strings Tension
AI Thread Summary
The discussion focuses on solving for the tension in strings AB and AE given a force of 600N exerted by AC. The angles involved include a right angle at BAC and a 37° angle at AEF, leading to the complementary angle of 53° for HAE. To find the tensions, it's suggested to resolve the forces into horizontal and vertical components and apply the equilibrium conditions, ensuring that the sum of forces in both directions equals zero. The confusion arises around identifying which components correspond to horizontal and vertical forces. The approach emphasizes breaking down forces rather than using non-perpendicular components for clarity in calculations.
max0005
Messages
50
Reaction score
0

Homework Statement




Considering the following diagram:

[PLAIN]http://img545.imageshack.us/img545/6301/physx1.jpg

AC is a force being exerted on chords AB and AE of magnitude 600N. Chord AE is attached to to surface FD. Chord B is attached to a second surface of which no information is given.

Angle BAC is a right angle.

Angle AEF is 37°.

AC is parallel to FD.

Find the tension of AC onto the system. (Ie: On each cable.)


Homework Equations





The Attempt at a Solution



If we elongate AF and AC such that:

[PLAIN]http://img232.imageshack.us/img232/2485/physx2.png

We obtain that angle FEA is equal to angle HAG. (37°)

As such, HAG's complementary angle HAE must be 53°.

If we accept HA to be equal to -600N (Opposite but equal to AC) and EA and BA to be its two components then we should be able to resolve it by stating that:

Horizontal Force = Resultant Vector * cos(Angle)
Vertical Force = Reulstant Vector * sin(Angle)

However, I am confused as how to go on from here? Which is the horizontal and which is the vertical vector?
 
Last edited by a moderator:
Physics news on Phys.org
I'm not sure of your angles, but i assume that AE makes a 37 degree angle with the horizontal.
It is best to solve these problems by breaking up all applied forces and tension forces into their vertical and horizontal components, then sum forces in the horiz and vert direction = 0 to solve for the unknown tensions. This is preferred rather than breaking up the applied force into non perpendicular components.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
7
Views
2K
Replies
4
Views
4K
Replies
25
Views
3K
Replies
15
Views
6K
Replies
3
Views
2K
Replies
3
Views
830
Back
Top