Static equilibrium: need help setting up the equation

  • #1

Homework Statement



A rigid wire frame is formed in a right triangle, and set in a vertical plane as shown. Two beads, of weight W1 = 1.5 N and W2 = 2.5 N, slide without friction on the wires, and are connected by a cord. When the system is in static equilibrium, what is the tension in the cord, and what angle does it make with the first wire?[/B]


media%2F917%2F91704cae-0f69-4ed3-a7cd-f67f36e0d5b8%2FphpDDMkIA.png



Homework Equations




The Attempt at a Solution

 

Answers and Replies

  • #2
gneill
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Hi xxglamisxx. You must show a solution attempt before any help can be given.

What have you tried? What equations have you written? What principles do think are involved?
 
  • #3
well I start with a FBD for each mass and and rotate the Cartesian coordinates so that the y axis is along the sides of the triangle for each mass and the x axis follows the normal force. I sum the x and y components to zero solving for either the normal force or tension force. I begin to have trouble because idk if there is a way to find theta before tension is found, but when i try to find tension, i have a bit of trouble relating the two masses to each other, ie: ending up with too many unknown (at least i think) variables.

when i try solving for T i end up with a sin(theta) or cos(theta) without having an actual angle, and i dont belive there is enough information given to use the arc trig functions to find theta.

any help is appreciated

thanks
 
  • #4
gneill
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If you end up with two equations with two unknowns then you should be doing alright.

Can you show us the equations that you've obtained from the FBD's?
 
  • #5
If you end up with two equations with two unknowns then you should be doing alright.

Can you show us the equations that you've obtained from the FBD's?

Sure.

Note: angle phi referenced below is the angle between horizontal and our tension cable on the right.

for M1 i have:

sum of forces in the x: N(sub1) = -G(sub1)cos(30) + T(sub1)sin(phi)

sum of forces in the y: T(sub1) = (G(sub1)cos(30)) / (sin(phi)


M2

sum of forces in the x: N(sub2) = G(sub2)cos(30) + T(sub2)cos(phi)

sum of forces in the y: T(sub2) = (-G(sub2)sin(30)) / (sin(phi)
 
  • #6
gneill
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I'm not sure why you've introduced the angle phi. Surely it would be best to stick with the given angle θ since that's what you're meant to find. But regardless, what have you chosen as your x and y axis directions? I'm trying to figure out where the 30° angle reference is coming from in your two sets of equations. Perhaps you can post a picture?
 
  • #7
let me see what i can do about the picture, as for the X and Y axis, each mass has its own. The one on the left, M1, has its Y axis along the left leg of the triangle, and M2 has its Y axis along the right leg of the triangle, the X's are perpendicular to their respective Y axis

phi was introduced, just as a place holder
 
  • #8
I'm not sure why you've introduced the angle phi. Surely it would be best to stick with the given angle θ since that's what you're meant to find. But regardless, what have you chosen as your x and y axis directions? I'm trying to figure out where the 30° angle reference is coming from in your two sets of equations. Perhaps you can post a picture?
going to start over on this problem fresh, but im currious...is there a way to find theta before
 
  • #9
I'm not sure why you've introduced the angle phi. Surely it would be best to stick with the given angle θ since that's what you're meant to find. But regardless, what have you chosen as your x and y axis directions? I'm trying to figure out where the 30° angle reference is coming from in your two sets of equations. Perhaps you can post a picture?
Ok so I re wrote the problem, with FBDs for each mass, and used θ instead of phi here is a picture of what i have:
WP_20160923_003.jpg
 
  • #10
haruspex
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For M1, you wrote that the gravitational force component in the X direction (normal to the wire here, right?) is Mg cos(30), the slope of the wire to the horizontal being 60 degrees. A thought experiment: if the slope had been 90 degrees, what would the normal component have been? What would cos(90-slope) give you?
 
  • #11
gneill
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Let me provide a diagram of the M1 situation that may be a bit easier to read.

upload_2016-9-23_18-53-13.png


Note that you're really only interested in the components of the forces that are directed along the wire. Forces normal to the wire can't do anything as there's no friction involved here and the wire prevents motion in that direction.
 
  • #12
For M1, you wrote that the gravitational force component in the X direction (normal to the wire here, right?) is Mg cos(30), the slope of the wire to the horizontal being 60 degrees. A thought experiment: if the slope had been 90 degrees, what would the normal component have been? What would cos(90-slope) give you?

youre right, that was a mistake on my part. to answer your question the normal force would be 0 degrees. for my x gravity component for m1 i changed it to cos 60 and respectively for the y component.

However, thanks to some guidance from @gneill i solved the problem.

method:

derived my formulas for T1 and T2, set them = to each other and used arctan to find the angle theta. from there i used my formula for t1 and plugged in theta to find the tension in the cable.

theta = 43.9, T = 1.80N
 

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