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Homework Help: Force vectors equilibrium and linear equations

  1. Mar 11, 2006 #1
    I have an example and a problem i am working on. The example is as follows: There are three strings. One hangs down with 100N while of the other two one goes up to the right from the equilibrium point at 45 degrees and the other goes left. I believe this set of Rx and Ry vector components is solving through the algebra method of linear equations. The string going up and to the right has an x and y component. The string going sideways only has an x component. Due to the simplicity of this problem it is possible to solve for the tension of the string going up by simply moving 100N to the other side and taking sin 30 and then moving it under 100N. This is put in the component of the horizontal string. This string is simply T1 cos 30. This gives us T3.

    The problem i am working on gives three string tensions and three weights. I solved for the first two weights and the first two tensions. The remainder of the problem is drawn as follows:

    An equilibrium point with a weight hanging on it of unknown value N. The right string is attached to the wall at 50 degrees and the left string goes up and over a pulley leaving from an angle of 30 degrees. It was already solved that the tension of the left string is 14.1 (there are 20 N coming from that side at equilibrium in another point).

    This is not a simple problem where you can just solve for one string at a time because you do not know the weight N. In the book example you could do this. I believe that the way to solve this problem is to set up one whole x component for all forces represented and then a y component for the whole thing. You then put one into the other.

    This gives:

    0 = Rx = 14.1N - T3Cos 50 - ?N
    0 = Ry = 14.1N - T3Sin 50 - 20N

    (This component system has 14.1 as T1 coming from the left and T3 as the unknown string going up to the right)

    I put 20N on the y axis because it is pulling on the left string (theres 20N in equilibrium coming from there) and 'unknown # Newtowns' on the x axis because it is hanging down from the equilibrium point which we are solving for.

    In the book example if was possible to isolate N and solve everything against it. Here we have two Newton values and we do not know one of them. I am trying to solve this like a system of linear equations by isolating for N and finding everything from it. I cannot find a way to merge the two N values however.

    What is the trick to solving for the sum N so that i can find all the other tensions from it? Is this the same type of problem where you solve for one T off of N and then plug it into the other or is this a different type of math? If it is a different type of math where do i look in lower level math or trig or calc to find it?

  2. jcsd
  3. Mar 11, 2006 #2
    The typical method that works is called the "substitution method." There are other techniques, of course, and depending on the problem some might work faster than others. I usually tell my students to learn this one as it's the simplest to implement (though admittedly, in certain problems the algebra can get hairy).

    Step 1:
    What you want to do is to solve one of your equations for one of your unknowns. When given a choice of which equation to solve and which unknown to solve for, my advice usually runs along the following lines:

    1) If it's quadratic, you don't usually want to pick this to solve first.
    2) Pick the easiest looking equation and variable to solve for.

    For example, in your system I would pick the unknown weight in your first equation, simply because it has no coefficient out in front of it.

    Step 2:
    Then you take that value of the unknown that you just solved for and plug it into your other equation. That now gives you one equation in one unknown and you can solve that. (Here's where the algebra might get a little nasty, particularly if your second equation is a quadratic.)

    Step 3:
    Take the solution to Step 2 and plug it into your equation from Step 1.

    For my garden variety student (that is to say, non-Physics major) I recommend learning one method to solve systems of equations and becoming a master at it. If you are serious about Physics, I would recommend reading up on some matrix techniques to solve systems of equations. (Look up "Gaussian elimination" in a book and you'll find them.) And if you are an engineer, learn how to solve them on your calculator...engineers usually don't care how they get the answer as long as they get them! :rolleyes:

  4. Mar 12, 2006 #3
    I am assuming that the following components are correct:
    0 = Rx = 14.1N - T3Cos 50 - ?N
    0 = Ry = 14.1N - T3Sin 50 - 20N
    If i do what you said and solve for ?N in Rx i will still have the unknown T3 and the unknown ?N. The book problem was easier because there was no y component so i believe T3 had a value. I won't have solved T3 or ?N. They will both be unknowns and equal one another.
    Also, does 20N go on the y axis and ?N go on the x, i'm not even sure about this.
  5. Mar 12, 2006 #4
    Okay. Four things.

    First, I hadn't really looked at your problem except from the standpoint of how to solve systems of equations, so I didn't see this:

    Second, it appears that you may simply solve your bottom equation for T3 since there is only one unknown in it. I didn't notice that before. Sorry.

    Third, I'm trying to reconstruct your problem and I'm having a difficulty.
    I've got you up to the point where you mention the 20 N. As I understand it there are three strings off your equilibrium point. One hangs directly downward (your unknown force N), one is acting up and to the right at 50 degrees. The last is acting upward and to the left at 30 degrees and you calculated to have a tension of 14.1 N. So where did the 20 N come from and what does it act on?

    Fourth, I recommend NOT putting units into your equations until you get to the final answer. Dimensional manipulation is a good thing to be good at, but all of the different symbols for units and quantities are easy to confuse in an equation.

  6. Mar 12, 2006 #5
    Thanks. Let me clarrify. The string going off to the left from the equilibrium point goes over a pully and comes back down. There is hits another equilibrium point. At this point a weight of 10N is hanging straight down. (The angle up to the pully is 45 degrees on this new side). From this equilibrium point another line goes to the left and goes over a pulley and straight down, hanging from it is the given weight of 10N. This last pulley is making a 90 degree angle.

    0 = Rx = 14.1N - T3Cos 50 - ?N
    0 = Ry = 14.1N - T3Sin 50 - 20N

    This is what i think the components are. I didn't know that you could combine 14.1N (which is a string tension T2) with the 20N from the block.
    Is this correct, you can combine the N values of the tension of the string (T) with the N value of the block? Something must be wrong with my components because i would get:

    Ry = T3 = -5.9 (Sin 50)

    If the number is negative (-5.9) something must be wrong with my components.

  7. Mar 13, 2006 #6
    So let me run down the diagram again because I'm still lost as to where this 20 N block is.
    From left to right:
    We have a 10 N block hanging straight down. A string (string 2) goes up from it to a pully. The string (2) goes over the pully and up and to the right at a 45 degree angle to another pully. The string (2) goes over that pully and goes down and to the right at a 30 degree angle to an equilibrium point. At the equilibrium point there are two other strings attached: one (string 3) goes up and to the right at a 50 degree angle and is attached to a wall; the other (string 1) goes straight down and is connected to a block of unknown weight.

    Let me know if that's right or correct me. Then we can work on applying Newton's Laws and get this sorted out.


    BTW Your original post said something about a 100 N weight? Is that a typo for the 10 N or is the 10 N a typo for the 100 N?
  8. Mar 13, 2006 #7
    Yes, you got the whole thing correct in your description except you missed one thing. Let
    me explain the diagram fully:

    (Note the names of the strings as the book is different then what we have labeled them as)

    A 10N block is hanging from T1 all the way out at the left. It goes up and over a pulley and straight out to the right. There there is an equilibrium point. A rope hangs down from this equilibrium point with a second 10N block. Leaving the equilibrium point and going up and at a 45 degree angle is T2. It goes over and comes down again. It hits another equilibrium point at 30 degrees. At this second equilibrium point a block hangs down of unknown weight in newtons. Going up from this equilibrium point and up and to the right is T3 with an angle of 50 degrees.

    There is no 20N block, there are two 10N blocks and i am combining them. I am not sure if they are both on the x axis. Last time i combined them and put them on the y axis. Now i think they both go on the x axis.

    As follows:

    0 = Rx = 14.1N - T3Cos 50 - 20N
    0 = Ry = 14.1N - T3Sin 50 - ?N

    Also, 100N was a typo. There are two 10N blocks.

    Last edited: Mar 13, 2006
  9. Mar 13, 2006 #8


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    There's nothing wrong with having a negative component - that just tells you which direction the vector is pointing. I don't really understand your word description, either, but I do understand what your equation describes.

    If left is negative on the x-axis and right is positive; and down is negative on the y-axis and up is positive; then:

    You have 20N up and to the right at a 45 degree angle (the Rx component is +14.1N and the Ry component is +14.1N). T3 is down and to the left 50 degrees below the horizontal (technically this 230 degrees counter clockwise of your positive x-axis). The last weight is 20N down with an unknown x component (which means putting a minus or negative sign is a bit presumptive, plus increases the chances of confusion by having a negative negative answer).

    To be honest, that doesn't sound like what you described in words.
    Last edited: Mar 13, 2006
  10. Mar 13, 2006 #9
    I revised my message with a better description of the problem in edit. If somebody could please check it over again. I think you can get the components from it more easily. It is message #7. Thanks.
  11. Mar 13, 2006 #10


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    Edit: disregard the comment about the 30 degrees - I don't know what I was thinking.

    In any event, at the first equilibrium point, one 10N block is pulling towards the left (neg x), one block is pulling straight down (neg y). T2 has to balance both of those components, so has 14.1N at 45 degrees.

    The magnitude stays constant, but you now need the angle relative to the second equilibrium point (30 degrees).

    The new horizontal component of T2 has to be balanced by T3 (T2 being negative and T3 being positive). Knowing the horizontal component, you can determine the vertical component (positive since it's angled upward at 50 degrees). The sum of the vertical component of T2 (positive y) plus the block of unknown weight (neg y since it hangs straight down) plus the vertical component of T3 (pos y) must equal zero.
    Last edited: Mar 13, 2006
  12. Mar 13, 2006 #11
    If i know that T2 is 14.1N do i need to create x and y components for it such as follows:

    T2x = Cos30 (20N)
    T2y = Sin30 (20N)

    The magnitude must be 20N since this is what is coming from that side. This seems redundant and is obviously incorrect. We know the tension of T2 is 14.1, it is not going to change based upon the angle and from which side you look at it.

    It would probably be simpler to just assume that T2 is 14.1N to represent the whole equation and put it against T3?

    I could rewrite the components with recalculated T values from new T2x and T2y coming from a new angle. These could be placed against T3x and T3y. This would be like saying:

    Rx= Recalculated T2 Value based on new angle - T3(cos50) - 20N
    Ry= Recalculated T2 Value based on new angle - T3(sin50) - xN?

    This is of course redundant and obviously incorrect. We already have them. It can't be that complicated. It should be my original components from earlier messages:

    0 = Rx = 14.1N - T3Cos 50 - 20N
    0 = Ry = 14.1N - T3Sin 50 - ?N

    Also, if i can just represent T2 as 14.1 and put it into the Rx and Ry components, is it (-14.1)?

    Then this would change everything to:

    0 = Rx = (-14.1N) - T3Cos 50 - 20N
    0 = Ry = (-14.1N) - T3Sin 50 - ?N

    Last edited: Mar 13, 2006
  13. Mar 13, 2006 #12


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    Your first problem is with the first equilibrium point. The two 10 N weights do not add up to 20 N. The weight hanging over the pulley on the edge exerts a force along the X-axis. (Along the pulley, the 10N force is always perpendicular to the pulley's radius, so the weight is converted to a horizontal force). The weight hanging straight down exerts a force along the Y-axis. The two forces are 90 degrees apart, so you have to use the Pythagorean Theorem to find the magnitude.

    Once you have the magnitude correct (14.1 N), you do have to break the 14.1 N force into its X and Y components. You can't just put the magnitude of T2 against the magnitude of T3 since the two forces are at different angles.

    At each equilibrium point, all of the forces along the X-axis should add up to zero and all of the forces along the Y-axis should add up to zero and each axis has to be handled separately.
  14. Mar 13, 2006 #13
    Okay, so following what was said in the other posts I am simply going to start the whole problem all over again. (I'm trying to follow at least some of your notation, but I'm likely to run all over some of it too. Sorry!) All of my coordinate systems have +x to the right and +y upward.

    Let's do equilibrium point 1, the one on the left.
    We've got a tension (T1) acting to the right. Doing a Newton's Law problem on the stationary hanging block of 10 N, we find that T1 = 10 N. We've also got a tension (T4) hanging straight down. This is also 10 N from the 10 N block on this string. Finally, we've got a tension (T2) acting up and to the right at an angle of 45 degrees.

    [tex]\Sigma F_x=-T_1+T_2cos(45) = 0[/tex]
    We could do the y equation, but it gets us nothing we don't already know from this. So we can get T2 from this equation (T2=14.1421 N as you already had in your original post. However, it seems you had derived it incorrectly, so I thought I'd post the method.)

    Now, the second equilibrium point. We've got a tension (T2) acting up and to the left at 30 degrees. We know the value of this from the above paragraph. We've got a tension (T5) hanging straight down. The value of T5 will be the size of your unknown hanging mass, which I'll call N (the symbol the original post gave for it. Personally I'd prefer W or something, but there you are.) We've got a tension (T3) up and to the right at 50 degrees.

    [tex]\Sigma F_x=-T_2cos(30)+T_3cos(50)=0[/tex]
    [tex]\Sigma F_y=T_2sin(30)+T_3sin(50)-N=0[/tex]

    We can use the x equation to solve for T3, then use that value in the y equation to find N. (I got N = 28.7 N or so if I did it right.)

  15. Mar 13, 2006 #14
    For the following components i have 2 questions:

    Fx = -T2cos(30)+T3cos(50) = 0
    Fy = T2sin(30)+T3sin(50)-N = 0

    First, in Fx, T2 is negative. In Fy, T2 is positive. Further, in Fx and Fy both T3 are positive. Is this because the y components are parallel lines to one another and are not considered opposites? In the x components the motion is in opposition so one of the T2 is negative? What is the reason for the Fx T2 to be negative?

    Second, the book answer for the tension of the string going up at 50 degrees is 19.1N. I am getting 18.99N. When i plug it into Fy i get the block to be 21.59N. The book says 21.7N. I am getting a rounding error but you got 28.7N for the weight of the block. What is the convention of rounding a cosine value and also rounding a figure such as magnitude? Thanks alot.
    Last edited: Mar 13, 2006
  16. Mar 13, 2006 #15
    Could T2 in Fx be negative because it is considered to be in the trigonometric quadrant 2 making cosine a negative if the angle is greater than 90 degrees? You would have to consider T3 in quadrant 1 and the weight on the Y axis. Cosine and sine do not come into play for the weight since it does not have components. Am i correct to assume that the equilibrium point is the 0,0 of a rectangular coordinate system?
  17. Mar 14, 2006 #16
    The equilibrium point is usually considered as an origin. Basically you can think of the signs in two (equivalent) ways:
    1. The angle of T2 is 30 degrees above the -x axis. So the x component of the vector is "negative."

    2. The angle T2 makes with the +x axis is 150 degrees. cos(150)=-cos(30), so the T2 x component is negative.

    In none of the above are we calling T2 itself negative, just its component in one direction.

    As for rounding, if I am going to calculate a number in the middle of a calculation I usually keep 4 or 5 digits beyond what I'm going to round to at the end of the problem. If I'm using my wonderful TI-92 I can keep around 10 digits. That usually solves the rounding problem.

    Usually I avoid calculating a number in the middle of a calculation. Whenever I can I plug in the exact expressions I get into the final equation. This significantly reduces rounding errors. For example, using the first equilibrium point I got
    From the x-equation on the second equilibrium point I got, using this expression for T2:
    So N turned out to be:

    It looks ugly, but it's not that bad to type in if you've got a programmable type calculator.

  18. Mar 14, 2006 #17
    Backtracking: I have a question. The first given weight at the very beginning of the problem is hanging down and has a weight of 10N. It goes up and over a pulley and makes a right angle and goes straight to the right. At this point it hits an equilibrium point where a block is hanging straight down from the equilibrium point of unknown weight. A line goes up and to the right from this equilibrium point at an angle of 45 degrees. If we don't know the tension of the line going up and to to the right at 45 degrees, and we are looking for the weight of the second block, it was stated that we should use the pythagorean theorem in an earlier post.

    The tension of the string holding the first given weight is 10N, but this is not important to the problem. If we are using the pythagorean theorem i get this:

    sqr root(10squared + ysquared) = a

    10 is the value of the first block. Y is the value of the second block. 'a' would be the answer which is the second block in newtons.

    This does not give me the answer 10 which is the answer. It gives me:

    sqr root(10sqr + ysqr) = a
    100 + ysqr = sqroot(a)
    ysqr = sqroot(a) - 100

    This is close to 10 except that sqroot(a) is stuck in there.

    What is the geometric relationship between the tension of the string which is 10N going up and over to the right when it hits the equilibrium point? There is an imaginary hypotenseuse drawn here from the first block to the equilibrium point. Of course, this probably does not concern the second block which is on a second line, not the triangle. Thanks.

    "Because your one in a million" -SF 1987

    "There's no time for this" -TN 1984
    Last edited: Mar 14, 2006
  19. Mar 14, 2006 #18
    There is an error in my last message. The statement should be a(sqr) + b(sqr) = c(sqr). This alters the math to the following:

    10(sqr) + b(sqr) = c(sqr)
    100 = c(sqr) - b(sqr)
    10 = c - b

    Again, the answer is ten, but i have an extra variable.

    "Cross my heart" -M 1988
  20. Mar 15, 2006 #19
    The above quote is from BobG on an earlier post in this thread. What he's talking about is that at your equilibrium point there are two forces acting at right angles. You can get the magnitude of the force that puts the system into equilibrium by using the Pythagorean Theorem. ONLY when these forces are at right angles.

    Translating this into LaTeX:

    You squared one side of the equation and took the square root of the other.

    If we don't know the tension and we don't know the other weight, we can't solve the problem because there are too many unknowns. Fortunately, according to your problem we DO know the weight..it is also a 10 N block. So we've got 10 N acting to the right (from the first 10 N block) and 10 N acting downward (from the second 10 N block) so:
    which is the value for this force that I gave you before.

  21. Mar 15, 2006 #20
    I see that a(sqr) + b(sqr) = c(sqr) is for finding the tension of T2. This shouldn't work if the angle is anything but 45 degrees. If we altered the weights it still shouldn't make a difference if the line angles are making a right triangle. I thought the pythagorean thereom was for finding the weight of the second block if you do not know from the weight of the first.

    Question: What if we didn't know the second blocks weight or the tension of the line? What if we only knew the weight of the first block which is 10N? How do we get from the first block with a weight of 10N to the second block which is unknown? In it's original form the problem does not provide the answer that the second block has a weight of 10N. This is in the solution. It should be from these two blocks that we can find 14.1N which is T2. Assume that we do not know the second block or the tension of T2. How do we find the weight of the second block with only the firsts? Thanks.
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