Three pulling strings with tension

[rdn98]

I've included a picture.

Three strings, in the horizontal plane, meet in a knot and are pulled with three forces such that the knot is held stationary. The tension in string 1 is T1 = 3.6 N. The angle between strings 1 and 2 is 130° and the angle between strings 1 and 3 120° with string 3 below string 1 as shown.

a) Find the tension in string 2.
b) Find the tension in string 3.

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A mass of 2.1 kg is now placed on the knot and supported by a frictionless table in the plane of the strings.
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c) Find the acceleration of the mass if all the forces remain the same as above.

d) If the sizes and directions of T2 and T3 remain the same, but T1 is increased by 1.5 N, what is the acceleration of the mass?

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Okay. I'm having trouble finding the tensions of string 2 and 3. I did split up both vectors into their components, but I tried out some trig and its not working for me.

For string 2, I tried T2= 3.6/(cos 50). That doesn't work.
For string 3, I tried T3= 3.6/(cos 240). That didn't work either.

From the hints, I'm supposed to have two equations and two unknowns, but I don't know how to derive these equations.

If I can do these two, then hopefully I can do the rest of the problem.

 

[gnome]

There's those damned sines & cosines again.

Start by taking your diagram & drawing x and y axes with the origin at the attachment point of the 3 ropes.

Next, you have to resolve T2 and T3 into their x and y components. T1, of course, acts only along the x axis, so it has no y component.

Then you set up your equations.
one is:
the sum of all of the x components = 0
the other is:
the sum of all of the y components = 0

Hint:
each of the components will be a tension multiplied (not divided) by either the sine or the cosine of the appropriate angle.

You HAVE to learn how to do this on your own. Are you up for it?

Tell me, what are the x and y components of T2 and T3?

 

[rdn98]

Um, after a bit of work, I finally got these equations.

T2x = T2*cos(130)
T2y = T2*sin(130)

T3x= T3*cos(-120)
T3y= T3*sin(-120)

 

[rdn98]

Ok, I finally figured out the problem gnome. It was a lot of work, but thanks for the initial start.

 

[gnome]

Well, since you got the equations right, I guess you understand the concept now, so it was worth whatever amount of work you put into it.

 

[HallsofIvy]

Okay, since you have the solution now, here how I would do the problem.

First, set up a coordinate system so that string 1, with force 3.6 Newtons, is directed along the x-axis. (I strongly recommend writing down precisely how you select your coordinate system- it clarifies your thinking.) The force vector for the force imposed by string 1 is (3.6, 0). (I chose that coordinate system just to make that easy.)

String 2 is then at angle 130 degrees (counter-clockwise) to the positive x-axis and, thus, the force vector is (T2 cos(130), T2 sin(130)), just as you got.

String 3 is then at angle 120 degrees (clockwise) to the positive x-axis and, thus, the force vector is (T3 cos(-120),T3 sin(-120))= (T3 cos(120),-T3 sin(120)). Again, exactly what you have.

Since the knot does not move the total force is 0:
(3.6+ T2 cos(130)+ T3 cos(120), T2 sin(130)- T3 cos(120))= (0, 0) which gives the two equations
3.6 + T2 cos(130)+ T3 cos(120)= 0 and
T2 sin(130)- T3 sin(120)= 0
which you can solve for T2 and T3.