Solving a Simple Tension Problem: Finding the Tension in Perpendicular Cables

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In summary: That means that the sum of the components of ai+bj is zero. In summary, the problem involves finding the tension in two cables suspending a 125 lb weight from the ceiling. The solution involves using vectors and breaking down the forces into horizontal and vertical components. The sum of the components must equal zero for the weight to not accelerate. This requires solving two equations for the two unknown tensions. The lack of understanding may be due to a lack of basic vector knowledge.
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Homework Statement



A 125 lb weight is suspended from the ceiling by means of two perpendicular flexible cables. Find the tension (in pounds) in each cable. There is a picture attached. I'm new at posting so I'll try my best to describe this simple image. There is a line that runs horizontal representing the ceiling. On the left side of the ceiling, there is a line that runs South east at a 55 degree angle ending at the 125lb weight. On the right side of the ceiling there is a line that runs south west at a 40 degree angle ending at the 125lb weight. That is about it. The ceiling and the cables form a triangle.

Homework Equations


None given


The Attempt at a Solution


This is a problem in the vector section of my multivariable calculus book. The solution to a similar problem is given by defining the cables (arbitrarily called T1 and T2) as T1=(T1 cos40)i +(T1sin40)j and T2=(-T2cos55)i +(T2sin55)j. From there the solution states that T1 + T2 + F(force or -125j) = 0. And they solve algebraically. I am having trouble with the concept. I do not want to plug in numbers without knowing why. I would like to know why we use cosine and sine. (I have a vague idea, but not a firm understanding). I would also like to know how to solve the problem. If anyone has an idea, it would be appreciated. I feel as though this is a very basic and simple problem and I would like to get some good insight to it. Thanks in advance.
 
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  • #2
You use sine and cosine to split the forces at the specified angles into i and j components. Then you add all the vectors. The result must be zero because the suspended weight is not accelerating. If the sum of the vectors is zero, then the sum of the i components and the sum of the j components must both be zero. That gives you two equations in two unknowns for the magnitudes of the tensions. Does that help?
 
  • #3
This should probably be in the introductory physics homework help subforum.
 
  • #4
Thank you Dick and AUMathTutor for responding. I believe I need more information on vectors in the general sense. Dick, your answer makes sense to me. I believe that I understand the idea of separating the i and the j components. However it occurs to me that I am missing a very rudimentary understanding of vectors because I do not fully understand your statement: "If the sum of the vectors is zero, then the sum of the i components and the sum of the j components must both be zero." It is not that this statement is overly complex or that it is worded in a way that is hard to comprehend. It is that I cannot state with complete accuracy that I understand completely what you are saying. I understand the idea that since the weight is not accelerating that the force should equal 0. However I do not understand, in depth, that the the i components and the j components should both be zero.

I believe this to be a result of my own lack of knowledge on the subject. I am missing some key vocabulary. Therefore, I would like to change the aim of this query to pointing me in the correct direction of basic vector knowledge. Thank you again Dick and AUMathTutor for taking the time to respond. I believe that my lack of knowledge is preventing me from understanding the solution. Also, sorry for posting this in the incorrect subject.
 
  • #5
The force from each cable on the weight is a vector pointing in the direction of the cable. You can resolve that into [itex]\vec{i}[/itex] and [itex]\vec{j}[/itex] components (horizontal and vertical). For example, if one wire makes angle [itex]\theta[/itex] with the horizontal and has tension T, we can draw vertical horizontal and vertical lines making a right triangle with the wire as hypotenuse. In terms of force, that hypotenuse has "length" T so the horizontal and vertical "lengths" (forces) are [itex]T cos(\theta)[/itex] and [itex]T sin(\theta)[/itex] respectively. Since the weight is not moving, the total force on it must be 0 and that means the horizontal forces must sum to 0 while the vertical force due to the wires must sum to the weight of the object. That gives two equations to solve for the two tensions.
 
  • #6
PhysicsMark said:
However it occurs to me that I am missing a very rudimentary understanding of vectors because I do not fully understand your statement: "If the sum of the vectors is zero, then the sum of the i components and the sum of the j components must both be zero." It is not that this statement is overly complex or that it is worded in a way that is hard to comprehend. It is that I cannot state with complete accuracy that I understand completely what you are saying. I understand the idea that since the weight is not accelerating that the force should equal 0. However I do not understand, in depth, that the the i components and the j components should both be zero.

If ai+bj=ci+dj and if i and j were ordinary numbers then you wouldn't be able to tell me much about the relation between a, b, c and d. But they aren't, i and j are vectors representing directions that are at right angles to each other. The only way that equality can be true is if a=c and b=d. The up-down components have to be equal AND the left-right components have to be equal. Can you picture why that has to be? (In technical language, i and j are linearly independent.) Similarly, if ai+bj=0, then a=0 and b=0.
 

1. What is a simple tension problem?

A simple tension problem is a physics problem that involves determining the tension force in a rope or cable that is holding an object in place. It is a common type of problem in introductory physics courses.

2. How do you solve a simple tension problem?

To solve a simple tension problem, you need to first draw a free body diagram, which shows all the forces acting on the object. Next, you can apply Newton's second law (F=ma) to the object in the direction of the motion. Finally, you can use the equation for tension (T=ma) to calculate the tension force in the rope or cable.

3. What are some real-world applications of simple tension problems?

Simple tension problems have many real-world applications, such as determining the tension force in a bridge or suspension cables, calculating the force needed to lift an object with a pulley system, or determining the tension in a bungee cord during a jump.

4. What factors can affect the tension force in a simple tension problem?

The tension force in a simple tension problem can be affected by the mass of the object, the acceleration of the object, the angle of the rope or cable, and any other external forces acting on the object.

5. How can I check if my solution to a simple tension problem is correct?

To check if your solution to a simple tension problem is correct, you can use the equations of motion to make sure all the forces are balanced. You can also plug your solution back into the original problem to see if it satisfies the given conditions.

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