How Do I Solve These Equilibrium Equations with T₂ and T₃?

[Suarden]

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

 

[gneill]

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?

 

[SteamKing]

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.

 

[Suarden]

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

 

[Suarden]

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

 

[gneill]

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.

 

[Suarden]

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

 

[gneill]

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.

 

[Suarden]

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.

 

[gneill]

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.

 

[Suarden]

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:(

 

[gneill]

cos(30°) and sin(30°) both have simple numerical values. Replace them accordingly and multiply out the terms. Then transpose.

 

[Suarden]

Please help me I am still confused

 

[Suarden]

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

 

[gneill]

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.

 

[Suarden]

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

 

[gneill]

(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!

 

[wifi]

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.

 

[Suarden]

(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!

 

[Suarden]

Thank you so much gneill! :D

 

[gneill]

You're welcome. I'm happy to help!