Calculating Cable Length and Stretch for Uniform Iron Beam in Static Equilibrium

Click For Summary
SUMMARY

The discussion focuses on calculating the minimum length of a cable and its stretch when supporting a uniform iron beam weighing 254 kg and measuring 3 m in length, under static equilibrium conditions. The maximum tension the cable can withstand is 1300 N. Participants outline the equations for static equilibrium, including force and torque summations, and provide a method to derive the cable length using trigonometric relationships. Additionally, they suggest applying Hooke's law to determine the cable's stretch based on its material properties and dimensions.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Knowledge of trigonometric functions in physics
  • Familiarity with Hooke's law and Young's modulus
  • Basic mechanics of materials, including stress and strain
NEXT STEPS
  • Calculate the minimum cable length using the derived equations for tension and angle.
  • Research the properties of steel, including Young's modulus, for accurate stretch calculations.
  • Explore advanced static equilibrium problems involving multiple forces and torques.
  • Learn about the implications of maximum tension in engineering design and safety factors.
USEFUL FOR

Mechanical engineers, physics students, and anyone involved in structural analysis or materials science will benefit from this discussion, particularly those focusing on static equilibrium and material properties under load.

Juntao
Messages
45
Reaction score
0
Picture is attached.

The above figure shows a uniform iron beam of mass 254 kg and length L = 3 m. The cable holding the beam in place can take a tension of 1300 N before it breaks. (You may ignore the small mass of the cable in this calculation.)

======
a)What minimum length of the cable?
b) Assume the cable is made of steel and has a diameter of 1". How much will it stretch before it breaks?

============
Okay. I am stuck on part a so far. I know that this is a static equilibrium, therefore, the summation of all the forces in x and y direction equal zero.
Also, the summation of torques must also equal zero.

But I'm not sure how to write out the equations. Can someone give me a lending hand?
 

Attachments

  • cable.gif
    cable.gif
    2 KB · Views: 639
Physics news on Phys.org
Let me try to label some forces without drawing a picture:

The weight of the bar, W, acts at the center of mass of the bar (1.5 m from the left end (set your origin at the left end)).

The tension in the cable, T, has components Tx (clearly pointing to the left) and Ty (pointing up).

The mounting of the bar (pin joint) can support a vertical force, Ry (pointing up) and a horizontal force, Rx (pointing to the right).

Some equations for static equilibrium:

[tex]\Sigma F_y = R_y + T_y - W = 0[/tex]

[tex]\Sigma F_x = R_x - T_x = 0[/tex]

[tex]\Sigma M_o = T_y\cdot L - \frac{WL}{2} = 0[/tex]

You can use these to find the y-component of the tension (actually, you only need the moment equation), but not much else.

Say you've solved for Ty and you want to find the minimum length of cable. You know that for the minimum length of cable, the cable tension will be at its maximum (not a great plan from an engineering standpoint, but nonetheless...). So:

[tex]T = \sqrt{T_x^2+T_y^2} = 1300[/tex]

use that to solve for Tx.

The length of the cable is given by:

[tex]L_c = \frac{L}{\cos \theta}[/tex]

where Lc is the length of the cable, L is the length of the bar (3 m), and θ is the angle between the cable and the bar, so that:

[tex]\tan\theta = \frac{T_y}{T_x}[/tex]

That should allow you to find Lc

For part b, you should be able to use Hooke's law, σ = εE, to find the length it stretches (you have the tension, the cross sectional area, the unstretched length, and you can look up the Young's modulus).
 
I'm not sure as I haven't actually worked it through, but because both θ and Lc are unknown, you might also have to use
Ty = Ry

(This must be true since otherwise the bar would rotate about its center of mass.)
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 9 ·
Replies
9
Views
6K
Replies
6
Views
6K
Replies
12
Views
4K
Replies
4
Views
5K