- #1

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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

- Thread starter Suarden
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- #1

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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

- #2

gneill

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Two equations in two unknowns... everything else is constants. Have you studied how to solve simultaneous equations?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

- #3

SteamKing

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Science Advisor

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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

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Yeah, I am just confused with this >.< I have a problem in isolating the 100N when I substituted the value. Please help me Y_YTwo equations in two unknowns... everything else is constants. Have you studied how to solve simultaneous equations?

- #5

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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

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.

[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

gneill

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

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

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this is where i get stucked---> [T₃Cos60°+100N)/Cos30° x Sin30°] + T₃Sin60°=9810NSorry, 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.

- #8

gneill

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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

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[T₃Cos60°+100N)/Cos30° x Sin30°] + T₃Sin60°=9810NShow and explain the algebraic steps you used to get to that point.

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

gneill

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Okay, I see how you got there.this is where i get stucked---> [T₃Cos60°+100N)/Cos30° x Sin30°] + T₃Sin60°=9810N

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

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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:(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.

- #12

gneill

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- #13

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Please help me I am still confused

- #14

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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

gneill

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After that, all the terms will be separate and you can transpose as you wish.

- #16

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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

After that, all the terms will be separate and you can transpose as you wish.

- #17

gneill

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The terms are easily separated, and even easier to deal with if you replace the sines and cosines with their numerical values!

- #18

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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.

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.

Last edited:

- #19

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Thank you so much!!!!!!!!!!!!!!!!!!!!

The terms are easily separated, and even easier to deal with if you replace the sines and cosines with their numerical values!

- #20

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Thank you so much gneill!!!!!! :D

- #21

gneill

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

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