Calc Tension of 2 Angles: 590-N Ball Suspended from Cords

In summary, the problem involves finding the tensions in the horizontal and angled cords suspending a 590-N ball. The sum of horizontal and vertical forces on the knot must equal zero, and the horizontal component of T2 must be equal to T1. Using the angle of T2 and the trigonometric relationship between its components, T2 can be solved for and then used to find T1.
  • #1
neutron star
78
1

Homework Statement


A ball weighing 590-N is shown suspended from a system of cords. What are the tensions in the horizontal and angled cords?
http://img207.imageshack.us/img207/2276/4figure39alt.gif

[tex]T_{50}=___[/tex]
[tex]T_{90}=___[/tex]

Homework Equations


The Attempt at a Solution


[tex]T_{1}cos(90)-T_{2}cos(50)=590[/tex]
This problem is confusing me because of the 90 degree angle.
 
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  • #2
neutron star said:

The Attempt at a Solution


[tex]T_{1}cos(90)-T_{2}cos(50)=590[/tex]
This problem is confusing me because of the 90 degree angle.

T1 acts in the x-direction already. So your "T1cos(90)" is really just supposed to be T1. Do you understand?
 
  • #3
So, [tex]T_{1} - T_{2}cos(50)=590[/tex]? Still, I don't get how to use that equation to find either [tex]T_{1} or T_{2}[/tex]. You can divide by cos(50) but that won't help.

[tex]T_{2}cos(50)=T_{1}-590[/tex]...
or
[tex]T_{1}=T_{2}cos(50)+590[/tex]
 
  • #4
neutron star said:
So, [tex]T_{1} - T_{2}cos(50)=590[/tex]? Still, I don't get how to use that equation to find either [tex]T_{1} or T_{2}[/tex]. You can divide by cos(50) but that won't help.

[tex]T_{2}cos(50)=T_{1}-590[/tex]...
or
[tex]T_{1}=T_{2}cos(50)+590[/tex]

Actually this is wrong, the vertical component is sin(50). The sum of the vertical components is zero and the sum of the horizontal component is zero.

So your equation should be T1-T2cos(50)=0

Now consider what the the vertical component of T2 should be equal to.
 
  • #5
Wait, ok so

[tex]T_{2}=T_{1}cos(50)[/tex]
so
[tex]T_{1}sin(90)+(T_{1}cos(50))sin(50)-F_{g}=0[/tex] right?
[tex]T_{1}=0.670F_{g}[/tex]
[tex]T_{1}=590/.670=880.597N[/tex]

This doesn't seem right, what did I do wrong?
 
  • #6
Alright, I'm going to give you a few big hints :)

I'll call T1 the horizontal cord

and T2 the cord at an angle.

Drawing a free body diagram @ the knot of all three cords. the forces i have... is a downward tension force from the weight that is equal to 590 N, a tension force T1 holding my knot in the horzontal direction, and a tension force T2 at an angle to my knot.

Now because i know that a=0 (everything is in equilibrium (not moving)) so because of Newton's First Law, i know the sum of all the forces on my knot is = 0. You remember this?

So now i deduce that the sum of the horizontal forces = 0, and the sum of the vertical forces = 0.

T1 is acting horizontally on the knot. I know that T1 does NOT support ANY of the weight of the object, it merely PREVENTS it from moving horizontally. so i know that T1 is equal to another force equal and opposite to prevent the cord from pulling the weight into the wall. But when i look opposite to T1, i don't see another force pulling along the horizontal, BUT we have a cord (T2) that is on an angle to the horizontal. So i know that T2 supports ALL of the weight of the object(because T1 supports none) + pulls against T1 at an equal amount.

So i know that the HORIZONTAL COMPONENT of T2 (ill call it [tex]T2_x[/tex]) is EQUAL to T1, because all the HORIZONTAL forces are equal to 0..

Therefore [tex]T1 + T2_x = 0 \therefore T1 = -T2_x[/tex] so T1 exerts an equal and opposite force of [tex]T2_x[/tex]

I also know i can solve for [tex]T2_x[/tex] because with the angle @ T2 50 degrees.

i know that [tex] sin 50 = \frac{T2_x}{T2} \therefore T2_x = T2 \cdot sin 50 [/tex]

But i don't know what T2 IS! so until i can find out T2, i can't find out T2x and until i find T2x i can't find t1.. get it? How do you find T2? What is [tex]T2_y[/tex] equal to? and how can you use that to get the values your looking for? Hope this helps..

Senjai
 

What is the formula for calculating the tension of two angles in a system?

The formula for calculating the tension of two angles in a system is T = mg/(cosθ1 + cosθ2), where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and θ1 and θ2 are the angles between the object and the cords.

How do the angles affect the tension in the system?

The angles in the system affect the tension by changing the direction of the force applied by the cords. As the angles change, the components of the force in the vertical and horizontal directions also change, resulting in a different tension in the system.

What is the significance of a 590-N ball being suspended from cords?

This setup allows for the study of different forces acting on the ball, such as tension and gravity. By measuring the tension in the cords, scientists can also determine the direction and magnitude of the force applied by the cords.

How can this experiment be applied in real-life situations?

This experiment can be applied in various real-life situations, such as determining the tension in a suspension bridge or the tension in cables supporting a heavy object. It can also be used in engineering and construction to ensure the stability and safety of structures.

What are some potential sources of error in this experiment?

Some potential sources of error in this experiment include the accuracy of the measurements, the effects of air resistance on the ball, and the assumption of idealized conditions (such as perfectly vertical and parallel cords). Human error in setting up the experiment and taking measurements can also contribute to inaccuracies.

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