Tension in two wires at unknown angles

In summary: I don't know. Halfway? 45 degrees? 60 degrees?In summary, an object at rest suspended from two strings (A and B) with A shorter than B pulls on point O with a force of 20 N. The force exerted by B on point O points downward and to the left, the magnitude of the force exerted by B on point O must be smaller than that of A, the magnitude of the vector sum of the forces exerted by A and B is equal to 20 N, the vector sum of the forces exerted by A and B points upward and to the right, and the magnitude of the force exerted by B must be less than or equal to 20 N
  • #1
MickeyBlue
26
2

Homework Statement


An object at rest is suspended from two strings (A and B) as shown in the diagram, with A shorter than B. The object pulls on the point O with a force of 20 N. Each of the strings also exerts a force on the Point O. The angle between the strings at O is 90o.

prob09.gif

Which of the statements below are true? (If A and E are true, and the others are false, enter TFFFT).

A) The force exerted by B on point O points downward and to the left.
B) The magnitude of the force exerted by B on point O must be smaller than that of A.
C) The magnitude of the vector sum of the forces exerted by A and B is greater than 20 N.
D) The vector sum of the forces exerted by A and B points upward and to the right.
E) The magnitude of the force exerted by B must be 20 N or less.

Homework Equations


F = ma

The Attempt at a Solution


I know that the net forces in the horizontal and vertical planes will add up to 0 respectively, and that double angles and trig can be used to narrow down the given data into two unknowns (one value for tension and the angle). I think that the final answer is False, False, False, True, True. I'm at a bit of a loss as to how to think about the problem further.
 
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  • #2
Perhaps you could give a reason for each answer.
(BTW I disagree with only one of your answers. One other answer I might say the opposite, but I can see how one could say T and F for it!)
 
  • #3
A) F - In component form, the forces will act up and to the right
B) F - I think considering cos of this force to the block will make it bigger than A
C) F - The vector sum should be equal to 20N to keep the system in equilibrium
D) F? - The forces in the horizontal should be opposite and equal, which means the vector sum should be up in the vertical
E) T - If B were greater than 20N, there would most likely be movement in this direction
 
  • #4
A) F - In component form, the forces will act up and to the right - force in B is up and right. I'm not sure where components come in?
B) F - I think considering cos of this force to the block will make it bigger than A - not clear to me. cos of which angle? how big is that angle?
C) F - The vector sum should be equal to 20N to keep the system in equilibrium - Agree
D) F? - The forces in the horizontal should be opposite and equal, which means the vector sum should be up in the vertical - Agree
E) T - If B were greater than 20N, there would most likely be movement in this direction - I don't follow your logic here. If both A and B were greater than 20N pulling mainly sideways, they'd cancel each other and only their smaller vertical components would add to 20N. But I do think B is less than 15N.
 
  • #5
Merlin3189 said:
A) F - In component form, the forces will act up and to the right - force in B is up and right. I'm not sure where components come in?
B) F - I think considering cos of this force to the block will make it bigger than A - not clear to me. cos of which angle? how big is that angle?
C) F - The vector sum should be equal to 20N to keep the system in equilibrium - Agree
D) F? - The forces in the horizontal should be opposite and equal, which means the vector sum should be up in the vertical - Agree
E) T - If B were greater than 20N, there would most likely be movement in this direction - I don't follow your logic here. If both A and B were greater than 20N pulling mainly sideways, they'd cancel each other and only their smaller vertical components would add to 20N. But I do think B is less than 15N.

  • A: where the resultant B can be resolved into vectors on the x- and y-axes
  • B: I'm sorry, I confused this with determining magnitude of work.
  • E: As far as this goes I know it can be said that Ay + By = 20N to counter the weight, regardless of which one is greater. If this refers to force in the vertical, then B would have to be less than 20N with A making up the difference. It's just the mathematical relationship between B and the acute angle that made me unsure. I understand it now.
Thank you for sacrificing some of your Saturday night!
 
  • #6
MickeyBlue said:
B: I'm sorry, I confused this with determining magnitude of work.
You need to be careful with B. I draw your attention to two facts:
  • You are not told angle alpha
  • The word "must" in the question
Write out the balance of forces equations for the horizontal and vertical directions.
MickeyBlue said:
E: As far as this goes I know it can be said that Ay + By = 20N to counter the weight, regardless of which one is greater. If this refers to force in the vertical, then B would have to be less than 20N with A making up the difference. It's just the mathematical relationship between B and the acute angle that made me unsure. I understand it now
I don't think you do.
To answer this question you need to use the given 90 degree angle.
 
  • #7
I note it says A is shorter than B. So the length of A is in the range 0 to slightly less than B.
 
  • #8
I think that makes statement B true.
 
  • #9
For statement E... I think force B would be a maximum when length A is a maximum. The max length of A is just less than length B. Eg the angle is 45 degrees.
 
  • #10
At the risk of giving too much help.. if A is less than B what does that mean for Tan(alpha) and hence alpha.
 
  • #11
CWatters said:
it says A is shorter than B
Yes, I missed that.
 
  • #12
I completely forgot about Pythagoras. Thanks for all the help.
 

FAQ: Tension in two wires at unknown angles

1. What is tension in two wires at unknown angles?

Tension in two wires at unknown angles refers to the amount of force or pull exerted on each of the wires when they are attached to an object at different angles. It is a measure of the strength of the wires and is important in determining their stability and ability to support the object they are attached to.

2. How is tension in two wires at unknown angles calculated?

The tension in two wires at unknown angles can be calculated using the law of cosines. This involves using the lengths of the wires and the angle between them to determine the tension in each wire. It is also possible to use trigonometric functions such as sine and cosine to calculate the tensions if the angles and lengths are known.

3. What factors affect the tension in two wires at unknown angles?

The tension in two wires at unknown angles is affected by several factors, including the weight of the object being supported, the length and material of the wires, and the angle at which the wires are attached to the object. Other factors such as external forces or environmental conditions may also affect the tension.

4. Why is tension in two wires at unknown angles important?

Tension in two wires at unknown angles is important because it determines the stability and safety of the object being supported. If the tension is too low, the wires may break or the object may fall. On the other hand, if the tension is too high, it may cause damage to the wires or the object. Understanding and calculating the tension is crucial in engineering and construction projects.

5. How can tension in two wires at unknown angles be measured?

There are several methods for measuring the tension in two wires at unknown angles. One common method is to use a tension meter, which measures the force exerted on the wires. Another method is to use strain gauges, which measure the deformation of the wires caused by the tension. Additionally, computer simulations and mathematical calculations can also be used to estimate the tension in two wires at unknown angles.

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