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Force Equalization on an Anchor

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data
    So I am a climber and I also love physics. I only have first year physics under my belt, and I do not know the concept behind the following question I often think about while setting up a climbing anchor.

    Attached is a diagram of two 2-point anchor systems. the red lines building the two sides of the anchor can be thought of as two lengths of rope, and 3rd length of rope attached to the vertex of the anchor goes to the climber (load). In situation (a) when the anchor is less than 120 degrees the load is shared between the two bolts, each seeing less than 1 x the load.. i.e if a 150 lb person is attached to the anchor each bolt will see maybe 75 pounds. therefore making it advantagous to use the 2-point system to reduce overall load on individual bolts.
    In situation (b), However; when the angle is greater than 120 degrees the force is actually multiplied at each bolt. so if you have a 150 lb climber (load) attached to the anchor, each bolt will actually see around 200+ lb, makeing it extremely dangerious to use this system.

    Why does the force multiply like that and decrease so much when the angle is decreased? with my first year physics all I know how to do is use the angles and the weight of the load to determine the tension on each leg of the anchor and hence the force on each bolt...the numbers work but i dont know WHY they are increaseing and decreaseing as the angle changes? is it acting as a sort of lever causing torque on the bolt.. can someone please explain this?
     

    Attached Files:

  2. jcsd
  3. May 18, 2010 #2

    PhanthomJay

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    Good point, but be careful with your bolt load calculations. It's due to joint equilibrium considerations per Newton's first law (sum of forces in vertical direction = 0, and sum of forces in horizontal direction = 0). The climbers weight is distributed equally (assuming the anchors are symmetrically located ) to each rope. This results in a vertical component of tension in each rope equal to half the climbers weight (T_v = 75 pounds, assuming a 150 pound climber). This is true regardless of the angle betwen the ropes. But when you look at the actual tension in the rope, which has both a vertical (T_v) and horizontal (T_h) component, the tension, T, in the rope is T_v divided by the cos of 1/2 the angle between the ropes (this follows from laws of equilibrium and geometry/trig/vector analysis). When that angle is 120 degrees, T = 75/cos 60 = 75/0.5 = 150 pounds . When that angle is 0 (bolts close to each other), T = 75/cos 0 = 75/1 = 75 pounds. When the angle is say 150 degrees, T= 75/cos75 = near 300 pounds. Please do not construe this as a recommendation for proper bolt locations when climbing. I'm not the rock climbing expert--- I defer to you on this, as I have the greatest respect for your skills.
     
    Last edited: May 18, 2010
  4. May 18, 2010 #3
    Thank you, Thanks for your reply.

    I can get the answers to the questions by breaking it into vectors as you did. however I still don't understand WHY this is happening (the force being multiplied as the angle increases) is it because like you say the vertical component will half the climbers weight no matter what the angle, but as you increase the angle, you increase the horiztontal component so in effect at around 180 degrees you would see 75 pounds on vertically downward and near 150 in the horizontal? im just trying to grasp the concept here? any thoughts?
    edit:
    upon doing calculations for 170 degree angle I still have 75 lb in the vertical component just as you said, but in the horizontal component I end up with 857.2 lb for a total force of 860.5 lb on the bolt. how is it possible to have 857.2 lb in the horizontal.. almost 6 times greater force than the original 150 lb. this logic is what baffels me
     
    Last edited: May 18, 2010
  5. May 18, 2010 #4

    PhanthomJay

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    Yes, it does seem like you are getting something for nothing, but note that when the bolts are close together, there is no horizontal component, since the ropes support the gravity load (weight) only. As you increase the angle to form the 'V' shape, each rope still must support the weight force, but each rope must also support the rope pulling horizontally on it from the other side. This is in order to satisfy the equilibrium equation in the horizontal direction, that forces in this direction must sum to zero, where the magnitude of such opposing forces are determined form the trig relationships of the force triangle. A good example of the high tension forces resulting for small lateral loads is the 'vehicle stuck in the mud' problem. By fastening a strong taut rope between the vehicle and a tree, you can apply a relatively small force at the mid point of the rope, and displace the rope laterally to form a wide angled 'V', and thus produce high tensile forces in the rope to get the vehicle to slightly move.
     
  6. May 18, 2010 #5
    Thanks very much for your explanation its starting to clear things up for me.
    and very cool truck in mud example. will possibly post another question regarding forces from climbing tomorrow. found a very cool one the today that confuses me quite a bit..
    I appreciate your help.
     
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