Physics question (EQUILIBRIUM) static?

In summary: Look at this example:Consider the equation ## a\cdot sin(90°) + b\cdot cos(0°) -5 = 0##. We know (or should know) that ## sin(90)° = 1 ## and ## cos(0°) = 1 ##.Substituting, this gives ## a+b -5 =0##. Thus, if we had another equation and wanted to eliminate say, ## a ##, we could solve for ## b ##. Doing so we find ## b=5-a ##.So say the equations were## 3a + 4b = 12 #### -2a + 5b = 6 ##Solving for ##
  • #1
Suarden
19
0
I already have this equations. But I'm confused on how to solve it, because of this 100.
ΣFx=0
T₂Cos30°-T₃Cos60°-100N=0

ΣFy=0
T₂Sin30°+T₃Sin60°-9810N=0

How will I solve this? Please I Really have a problem on simplifying it, I need to solve for the T₂ and T₃
 
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  • #2
Suarden said:
I already have this equations. But I'm confused on how to solve it, because of this 100.
ΣFx=0
T₂Cos30°-T₃Cos60°-100N=0

ΣFy=0
T₂Sin30°+T₃Sin60°-9810N=0

How will I solve this? Please I Really have a problem on simplifying it, I need to solve for the T₂ and T₃

Two equations in two unknowns... everything else is constants. Have you studied how to solve simultaneous equations?
 
  • #3
Evaluate the trig terms and substitute the numerical values into your two equations.
Move the constant terms (100N and 9810N) to the right hand side of the equations.
What is left? You have two equations in two unknowns (T2 and T3).
Solve for the unknowns. I'll leave the choice of method up to you.
 
  • #4
gneill said:
Two equations in two unknowns... everything else is constants. Have you studied how to solve simultaneous equations?
Yeah, I am just confused with this >.< I have a problem in isolating the 100N when I substituted the value. Please help me Y_Y
 
  • #5
SteamKing said:
Evaluate the trig terms and substitute the numerical values into your two equations.
Move the constant terms (100N and 9810N) to the right hand side of the equations.
What is left? You have two equations in two unknowns (T2 and T3).
Solve for the unknowns. I'll leave the choice of method up to you.
You cannot just transpose the 100N in the right side, I already tried that, when substituting the values, 100N is attached with the Sin30°, this is the resullt of substituting the values


[T₃Cos60°+100N)/Cos30° x Sin30°] + T₃Sin60°=9810N
^
^
How would I simplify/solve that equation? >.< I know I cannot just immediately transpose the 100N because it is attached in the first equation. Please helppppppppp
 
  • #6
Suarden said:
Yeah, I am just confused with this >.< I have a problem in isolating the 100N when I substituted the value. Please help me Y_Y

Sorry, what is ">.<" ? Some mathematical notation?

Start by replacing the trig terms with numerical values. Note that 30° and 60° have well known sines and cosines based upon the 30-60-90 (1-2-√3) triangle.

Show your attempt so that we may see where you're getting stuck or going awry.
 
  • #7
gneill said:
Sorry, what is ">.<" ? Some mathematical notation?

Start by replacing the trig terms with numerical values. Note that 30° and 60° have well known sines and cosines based upon the 30-60-90 (1-2-√3) triangle.

Show your attempt so that we may see where you're getting stuck or going awry.
this is where i get stucked---> [T₃Cos60°+100N)/Cos30° x Sin30°] + T₃Sin60°=9810N
 
  • #8
Suarden said:
this is where i get stucked---> [T₃Cos60°+100N)/Cos30° x Sin30°] + T₃Sin60°=9810N


Show and explain the algebraic steps you used to get to that point.
 
  • #9
gneill said:
Show and explain the algebraic steps you used to get to that point.
[T₃Cos60°+100N)/Cos30° x Sin30°] + T₃Sin60°=9810N

My problem is I cannot just transpose the 100N to the right side so That I could easily solve the equation, and if I multiply it with the Sin30° I won't be able to Isolate it completely because it will have a Symbol in Newtons.
 
  • #10
Suarden said:
this is where i get stucked---> [T₃Cos60°+100N)/Cos30° x Sin30°] + T₃Sin60°=9810N

Okay, I see how you got there.

You're on the right track. But replace the sines and cosines with numeric values before going further! sin(30°) is 1/2. Cos(30°) = √3/2. Do similarly for sin and cos of 60°. It'll make your life much easier if you remember these common trig values for common angles...they come up very often on problems!

You need to isolate T3 in your equation then substitute the result into the other equation to solve for T2.
 
  • #11
gneill said:
Okay, I see how you got there.

You're on the right track. But replace the sines and cosines with numeric values before going further! sin(30°) is 1/2. Cos(30°) = √3/2. Do similarly for sin and cos of 60°. It'll make your life much easier if you remember these common trig values for common angles...they come up very often on problems!

You need to isolate T3 in your equation then substitute the result into the other equation to solve for T2.
I'm still confused :( I really have a problem in Transposing the 100N to the right side. I can't just transpose it because it's in the parenthesis:(
 
  • #12
cos(30°) and sin(30°) both have simple numerical values. Replace them accordingly and multiply out the terms. Then transpose.
 
  • #13
Please help me I am still confused
 
  • #14
gneill said:
cos(30°) and sin(30°) both have simple numerical values. Replace them accordingly and multiply out the terms. Then transpose.
there's always a denominator that's why I cannot immediately transpose, this is the equation again: [(T₃Cos60°+100N)/Cos30°] x Sin30° + T₃Sin60°=9810N
 
  • #15
The way forward is to replace the sines and cosines with their numerical values, then multiply out the terms.

After that, all the terms will be separate and you can transpose as you wish.
 
  • #16
gneill said:
The way forward is to replace the sines and cosines with their numerical values, then multiply out the terms.

After that, all the terms will be separate and you can transpose as you wish.
I tried to do that but my problem is with the denominator, remember my numerator has this 100N. How will I be able to isolate my 100N? that is really my problem, even if I changed it to its numerical value and then simplify, there is still a denominator and the numerator which inclues the 100N as numerator :3
 
  • #17
(A + B)/C = A/C + B/C

The terms are easily separated, and even easier to deal with if you replace the sines and cosines with their numerical values!
 
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  • #18
Look at this example:

Consider the equation ## a\cdot sin(90°) + b\cdot cos(0°) -5 = 0##. We know (or should know) that ## sin(90)° = 1 ## and ## cos(0°) = 1 ##.

Substituting, this gives ## a+b -5 =0##. Thus, if we had another equation and wanted to eliminate say, ## a ##, we could solve for ## b ##. Doing so we find ## b=5-a ##.

So say the equation which we're trying to eliminate ## a ## is ##2a+b=2##. Substituting our value for b we find ## 2a+(5-a)=2 ##.

Simplifying we get ##a=-3##. Now that we have ##a##, finding ##b## is simply a matter of plugging the value of ## a ## into our previous equation. Doing so we find ##b=5-(-3)=8 ##.

As a check plug these into the original equation ##a+b-5=-3+8-5=-8+8=0##. Also, plugging these values into the other equation we get ##2a+b=2(-3)+8=-6+8=2##. These both agree, as they should.

Your problem is analogous, just messier values for sine and cosine.
 
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  • #19
gneill said:
(A + B)/C = A/C + B/C

The terms are easily separated, and even easier to deal with if you replace the sines and cosines with their numerical values!
Thank you so much!
 
  • #20
Thank you so much gneill! :D
 
  • #21
You're welcome. I'm happy to help!
 

1. What is static equilibrium in physics?

Static equilibrium in physics refers to a state where all forces acting on an object are balanced, resulting in no net force and no change in motion. This means that the object is either at rest or moving at a constant velocity.

2. How is static equilibrium different from dynamic equilibrium?

Static equilibrium is when an object is at rest or moving at a constant velocity, while dynamic equilibrium is when an object is moving with a constant acceleration. In static equilibrium, all forces are balanced, while in dynamic equilibrium, the net force is zero but the object is still in motion.

3. What is the principle of moments in static equilibrium?

The principle of moments states that in order for an object to be in static equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments. This means that the object must have a balanced torque in order to remain in static equilibrium.

4. How do you calculate the net torque in static equilibrium?

The net torque in static equilibrium can be calculated by multiplying the force applied to an object by the distance between the force and the pivot point. This can be represented by the equation Σm = 0, where Σm is the sum of the moments and must equal zero for static equilibrium to be achieved.

5. What are some real-life examples of static equilibrium?

Some real-life examples of static equilibrium include a book resting on a table, a person standing still, or a ladder leaning against a wall. In all of these situations, the forces acting on the object are balanced, resulting in static equilibrium.

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