Determine largest force when three ropes are joined at a point

[zeralda21]

Homework Statement

Three ropes, numbered 1-3, are joined at one point. Three persons pull in each separate rope, so that the rope is at rest. The angles between the ropes are shown in the figure (note that it is the lengths as illustrated, and not the size of the forces). Arrange the drag forces in the ropes, called F1, F2 and F3 respectively, according to their size (eg, "F1 <F2 <F3" or "F1 = F3 <F2").

http://bayimg.com/EaoNIaaDh

Homework Equations

No equations needed. Maybe some basic trigonometry.

The Attempt at a Solution

My thoughts:

First I conclude that the force depends on the length of the rope. If one rope is much shorter than the two other, then the force needed for the system to be at rest will be much larger for that rope. The same works in the opposite direction(large rope-->small force).

Since F_3 is significantly longer than F_1 and F_2, I conclude that both F_1 and F_2 are larger than F_3. Notice now that the length of rope 1 and 2 are equal. If forces F_1 and F_2 were equal, then the angle of F_3 with the horizontal axis would be 45 degrees which is not the case. The angle is actually less than 45 degrees and therefore is F_2 larger than F_1.

Hence: F_2>F_1>F_3. Correct answer: F_3>F_1>F_2

 

[tiny-tim]

hi zeralda21!

(try using the X₂ button just above the Reply box )

the length of the ropes has nothing to do with it

hint: draw a vector triangle ​

 

[zeralda21]

hi zeralda21!

(try using the X₂ button just above the Reply box )

the length of the ropes has nothing to do with it

hint: draw a vector triangle ​

Hi Tiny-tim :)

When drawing a vector triangle, I believe that I will need the magnitude and direction of a force. The direction is shown in the figure(along the rope of course) but how do I know the magnitude?

 

[tiny-tim]

d'oh!

that's the whole point of vector triangles …

you can still draw them without all the information …

in this case, you know the directions, so isn't there only one way the sides can join up?

 

[zeralda21]

d'oh!

that's the whole point of vector triangles …

you can still draw them without all the information …

in this case, you know the directions, so isn't there only one way the sides can join up?

Sorry about my incompetence of the subject tiny-tim :shy:

Yes, the forces can join in one way. And since F_3 is the hypotenuse of that triangle it means that F_3 is the largest force. But this right angled triangle can look different dependent on the magnitude of the forces. See picture for two examples:

http://bayimg.com/CAOpEaadh

How can I tell which one is larger?

 

[tiny-tim]

… But this right angled triangle can look different dependent on the magnitude of the forces. See picture for two examples:

http://bayimg.com/CAOpEaadh

How can I tell which one is larger?

not following you …

in your original picture http://bayimg.com/EaoNIaaDh the directions are fixed …

just slide the lines along until they join up! ​

 

[zeralda21]

not following you …

in your original picture http://bayimg.com/EaoNIaaDh the directions are fixed …

just slide the lines along until they join up! ​

I don´t really get it,sorry...The lines have lenghts (5, 5 ,7.5). I can´t see how it is possible to construct a triangle of these lenghts...

 

[tiny-tim]

The lines have lenghts (5, 5 ,7.5). I can´t see how it is possible to construct a triangle of these lenghts...

ah, no, the lengths are the lengths of the ropes …

the tensions could be anything

but the directions of the tensions are, obviously, the same as the (given) directions of the ropes!

does it make sense now? ​

 

[zeralda21]

ah, no, the lengths are the lengths of the ropes …

the tensions could be anything

but the directions of the tensions are, obviously, the same as the (given) directions of the ropes!

does it make sense now? ​

I´m not sure I am following good enough.. Correct me if I´m wrong:

*"just slide the lines along until they join up! ---> I need to construct a triangle where the lengths are the lengths of the rope(5,5,7.5) since the lines represents the lengths of the rope. I am stuck at this.

"tensions could be anything" OK.

"but the directions of the tensions are, obviously, the same as the (given) directions of the ropes! - Yes, I get this.

 

[tiny-tim]

*"just slide the lines along until they join up! ---> I need to construct a triangle where the lengths are the lengths of the rope(5,5,7.5) since the lines represents the lengths of the rope.

no

forget the lengths

you need to construct a triangle where the directions are the directions of the rope

since the lines represents the lengths of the rope and their directions,

and the triangle represents the forces of tension in the ropes

 

[zeralda21]

no

forget the lengths

you need to construct a triangle where the directions are the directions of the rope

since the lines represents the lengths of the rope and their directions,

and the triangle represents the forces of tension in the ropes

Alright, like this then : http://bayimg.com/haOPFaAdH .

If I follow "and the triangle represents the forces of tension in the ropes this:

In case ,my triangle is right, then the largest force is clearly, F_3, since it is a hypotenuse. But what about the other two vectors?

 

[zeralda21]

If I return to original picture I can form a smaller triangle that is equilateral to the "triangle of directions". For simplicity I chose lengths 3 (F_1), 4 (F_2) and hypotenuse 5 (F_3), therefore F_1 is larger than F_2. Is this OK?

 

[tiny-tim]

mmm … i'd be more convinced if you did it by altering the original diagram (http://bayimg.com/EaoNIaaDh ), so we can see how you did it

Alright, like this then : http://bayimg.com/haOPFaAdH .
In case ,my triangle is right, then the largest force is clearly, F_3, since it is a hypotenuse. But what about the other two vectors?

measure them!

(count squares)

EDIT: oh you did …

yes, it's 6 squares along the bottom and 4.5 along the side ​

 

[zeralda21]

mmm … i'd be more convinced if you did it by altering the original diagram (http://bayimg.com/EaoNIaaDh ), so we can see how you did it


measure them!

(count squares)

EDIT: oh you did …

yes, it's 6 squares along the bottom and 4.5 along the side ​

Well, there you go Thanks a lot tiny-tim!