Analyzing Forces in a Three-String Knot

  • Thread starter Thread starter CaptainSFS
  • Start date Start date
  • Tags Tags
    Strings Tension
AI Thread Summary
The discussion focuses on analyzing the forces acting on a knot where three strings are pulled, with tension in string 1 measured at 2.9 N. The angles between the strings are given as 130° and 120°, and the knot remains stationary, indicating that the net forces are zero. Participants emphasize the importance of using a free body diagram (FBD) to visualize the forces and develop equations based on the components of tension in the x and y directions. The key takeaway is to create two equations with two unknowns to solve for the tensions in the other strings. The conversation concludes with a participant successfully resolving their calculations after understanding the approach.
CaptainSFS
Messages
57
Reaction score
0

Homework Statement



showme.pl.gif


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 = 2.9 N. The angle between strings 1 and 2 is q12 = 130° and the angle between strings 1 and 3 is q13 = 120° with string 3 below string 1 as shown.

Homework Equations



F=ma

The Attempt at a Solution



I am unsure. I tried finding two equations with two unknowns and then tried solving for one, but I no idea what to do. my free body diagrams aren't helping either. Could someone explain how my FBD would look or at least explain what equations I should have come up with? Thanks for any help.
 
Physics news on Phys.org
CaptainSFS said:

Homework Statement



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 = 2.9 N. The angle between strings 1 and 2 is q12 = 130° and the angle between strings 1 and 3 is q13 = 120° with string 3 below string 1 as shown.
I am unsure. I tried finding two equations with two unknowns and then tried solving for one, but I no idea what to do. my free body diagrams aren't helping either. Could someone explain how my FBD would look or at least explain what equations I should have come up with? Thanks for any help.

The key phrase is the "knot is held stationary". That means that the net forces are = 0. It also means that the sum of the component forces in each direction is also 0 or it would move along that direction.

Now you can choose any coordinate system, but being lazy I like to choose at least one axis that makes things easier. Since the angles are given in terms of the relationship with T1, that would be my choice for x-axis.

Now develop equations for T1_x and T2_x and their angles and T3_x and since there is no Y component of T1 then another equation expressing T1_y and T2_y as a function of their angles.

2 equations. 2 unknowns. You don't need any more than that.
 
hey thanks! I worked out the rest of it just fine. :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top