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Homework Help: Force on anchor calculation help.

  1. Jul 23, 2008 #1


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    With reference to the attached picture (very basic).
    - Point X, Y & G are anchored.
    - The 100kg load is suspended by point G (like a 1:1 pulley)
    - Assume F is 90degs

    My observations:
    With an angle of 90deg I see that approx 70% of the load from point A would be transfered along the lines A-B , A-C.

    I also see that the load at point A would be close to double the actual load with the 1:1 setup.

    Now the line from X to Y needs to stay in approximately a horizontal line for the anchors X and Y to work.

    I see that the line D to E increases in tension once the load is applied at point A.

    Assume all lines have zero stretch and assume nil friction reducing the forces.

    Scenario 1: How much force is felt at points: A,B,C,D,E, X&Y?

    Scenario 2: If the load fell 1 meter what are the forces now with an increase in inertia?

    Really hope you can help clarify things here and I appreciate your time.

    Attached Files:

  2. jcsd
  3. Jul 24, 2008 #2


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    Welcome to PF!

    Hi BBB! Welcome to PF! :smile:

    Just go through the diagram from bottom to top …

    to start you off: what is the tension in AL (where L is the load)?

    then what is the tension in AG?

    then what is the tension in FC and FB? :smile:
  4. Jul 24, 2008 #3


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    Re: Welcome to PF!

    All right here is my guess as I really have very little education in this....
    There would be approximately 200kg felt at 'A'.

    With approx 70% transfered along A-B and A-C, I am guessing that there is about 140kg felt at the points D and E. So how much force (KN) is applied to the anchors X and Y by the the tension in line X to Y to stay in a semi horizontal plane when the force is loaded downwards on this X Y line?. (It needs to stay horizontal for the anchors X and Y to work) ?????

    Is it 140kg (1.4KN) felt at each anchor or 240kg? Or more?

  5. Jul 24, 2008 #4


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    Yes! :smile:

    But the way you should write it is this:

    Consider only the forces on L.
    There are only two … the weight and the tension in AL.

    So they must be equal and opposite. So the tension in AL is 100kg (or 980 newtons).

    There is no friction at A, and the rope GAL is continuous, so the tension in AG is the same.​
    hmm … you haven't actually said where you got 70% from. :frown:

    It should go something like this …

    Now consider the forces on A.

    There are only four … the tensions in AL AG AB and AC.

    Taking components horizontally, the tensions in AB and AC must be equal.

    Taking components vertically, the vertical components of the tensions in AB and AC must … ? :smile:
    Can you complete this?
  6. Jul 24, 2008 #5


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    The 70% is derived by cosines..
    100kg divided by 2 (2 lines AB and AC) over Cos (angle F divided by 2)

    Eg: 100/2 over cos(90/2) = 50/0.70 = 71 ie: 71% as 100kg was the load

    In my case substitute the load for 200 = 200 * 71% = 142. I said 140kg for ease of maths.

    So to answer my question; what is the KN force felt at X and Y? I need help / confirmation here. Edit to add that I dont see that the loads would be 140kg at X and Y because they are supporting the tension/200kg downward load by maintaining a horizontal line.
    Last edited: Jul 24, 2008
  7. Jul 24, 2008 #6


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    Hi BBB! :smile:
    ok, but you really need to write your answers out properly.
    Wait … you haven't worked out the tension in the rope DE yet.

    And if XD and YE are also ropes, then I agree, they can't be horizontal.
  8. Jul 24, 2008 #7


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    DE...... yes that is my question. Only basic tension is on this line to hold it 'horizontal' and keep the anchors X and Y in tack until the load is applied, then this line DE gets tighter in order to keep the line X Y horizontal, or at least as horizontal as possible.

    So when the load comes on, what tension is across D-E, X - Y and therefore how much does each anchor X and Y have to deal with.

    Yes they are all 'ropes' as such. So there can be flex from horizontal and that is what has to be stopped. Sorry should have pointed that out.

    A to B and A to C is a standard setup. And as I have figured out approx 142kg is felt where it attaches to the horizontal line XY at points B and point C ie the same points that line D and E are attached.

    If the line D E has no tension and is slack then the load would run on the 90 deg angle effectively to point X and Y through that conection point that B,C,D,E connect to. This wont work as X and Y need to oppose each others force and remain in as close to a horizontal plane.

    If we just look at the line X to Y and suspended 200kg directly from the center of the line X Y (DE), there would be downward deflection in line XY. If the defection was say only 10deg, then the angle formed where the load is suspended would be 170degs.

    Using 170 deg, the cosine is 574. So 200kg * 574% = 1148 kg felt at each anchor. So that means?? that you would need around 11KN of force on that line to make sure it only sags 10deg. So that is 11KN on each anchor or dived between??

    So how much force is needed between X and Y and therefore D and E to suspend the load when it is not actually suspended at the center of line X Y, ie: the force of 142kg is at points B and C...... Or should I be looking at the same way as I have calculated above, therefore meaning that when the load is applied at 'A' then there will be effectively force multiplication and a force at X and Y of about 11Kn or is it 5.5Kn??

    I am confused!!!

    Does this make better sense?

    Thanks for helping out.
    Last edited: Jul 24, 2008
  9. Jul 25, 2008 #8


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    Hi BBB! :smile:

    hmm … if XDEY was a solid beam, there would be no difficulty.

    Or if XDEY was a continuous rope, so that the tension was the same in all three sections.

    But there are three unknowns (two tensions and an angle), and only two available equations, so I don't see how you can find them without knowing something else, like the stretchability of the rope (yet the question specifies "Assume all lines have zero stretch"). :confused:

    I also don't understand:
    i] how can the load fall?

    ii] whatever does "an increase in inertia" mean? :confused: :confused:
  10. Jul 25, 2008 #9
    In regards to the original question. If you assume that the structure is pin jointed or tension cables then:

    With the system you have drawn the tension in member F-C and F-B will be 141.4kg (100/COS(45)) and the tension in member D-E will be 100kg (100/COS(45)).

    With the supports being horizontal at points X and Y the load is in theory is infinite. You need some vertical angle in the members or it will self destruct.

    If you were to drop the load the force generated would depend on the elasticity of the members.
    Last edited: Jul 25, 2008
  11. Jul 25, 2008 #10
    If members X and Y had 10 degrees of sag then the force on each anchor would be 567kg (100/TAN(10)).
  12. Jul 25, 2008 #11


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    Thanks for the replies Tiny Tim and Bolts.

    Lets ask the question this way then....

    If the line X-Y was continual and we load the center, we get sag and we can figure out the result.

    Now what happens when we suspend the same load but from two different points, like in my picture, which then come to the common point A? Does it place more pressure on the D-E to keep X - Y level, or does it place more or less force on the anchors X and Y when compared to having the force suspended from only one point in the center?

    To me D- E is what is feels the tension to hold X-Y level (or as level as possible) when the load is applied at 'A'. That in turn should then transfer to the anchors X and Y, by how much I don't know...

    Looking at the new attached pic. This represents in an exagerated form what the line between X - Y looks like when the load is on point A and it tries to pull the line from horizontal. Until a point in time where D-E takes up the tension to stop this downward force.

    This new angle at H is, say, 170deg (10 sag).
    Now I don't think the load at X and Y is simply 567 as the line D-E has equalising tension on it to stop this downward pull and I cant see how this is only 100kg (Cos45 etc). If it definitely is then that is good.

    SO is the answer simply that we have approx 567kg on each anchor X and Y and a tension of 100kg on line D-E?

    Bolts could you please explain your 567kg using TAN? I don't quite get it.


    Attached Files:

  13. Jul 25, 2008 #12
    If the sag in member x and y is 10deg from horizontal and the vertical component of that force is 100kg then the horizontal component will be 100/TAN(10).

    When I look at the new drawing I can see that the tension in members x and y is 575.9kg(100/SIN(10)). The horizontal component will be 567kg (100/TAN(10)) therefore the total tension in the member D-E will be 467kg (100/TAN(10)-100/TAN(45)).
    Last edited: Jul 25, 2008
  14. Jul 26, 2008 #13


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    Thanks bolts. Just to clarify the load of 100kg becomes 200kg due the 1:1 effect.
    So should I substitute 200 into you TAN equation? Using: (Load/2) / (COS)(170/2) I get 1147kg on each anchor. Not 567kg. And I am sure this formula has proved correct with lessor angles.

    Also could you please explain the theory on when and how you used SIN and also TAN.

    With D-E = 467kg. Why is 45deg used in the equation?

    Thanks again.
  15. Jul 26, 2008 #14
    The equations I presented are for each support. Given that the total load is 200kg then the sum of the vertical forces at the supports has to equal 200kg (100+100) and the vertical component on each support is therefore 100kg as use in my calculations. Sum of the vertical forces equals zero.

    Sin, tan and cos are use to calculate the vertical and horizontal force components in a tension member. I don't have enough time to explain in detail but it is done using basic trigonometry of the force vector try searching for "force vector component calculation".

    45 degrees is the angle of the member F-C and its horizontal component must be included so that the sum of the horizontal forces on pin C-E is zero. While on that topic you will notice that the sum of the vertical forces is at zero on this member also being 100kg in either direction.
    Last edited: Jul 26, 2008
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