formula for weight on two anchor points? [closed]
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What are the formulas for calculating the weight on X and Y in these diagrams assuming the angle A and load Z can change:
Y-hand & deviation images
physics
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closed as off-topic by John Douma, amWhy, Paul Frost, pre-kidney, John Bentin Jan 13 at 7:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Douma, amWhy, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
What are the formulas for calculating the weight on X and Y in these diagrams assuming the angle A and load Z can change:
Y-hand & deviation images
physics
$endgroup$
closed as off-topic by John Douma, amWhy, Paul Frost, pre-kidney, John Bentin Jan 13 at 7:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Douma, amWhy, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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What have you tried?
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– Larry
Jan 12 at 17:32
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Nothing, I don't know where to start. I just want a couple of formulas so I can transfer them into c++
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– cppVulkan
Jan 12 at 18:15
add a comment |
$begingroup$
What are the formulas for calculating the weight on X and Y in these diagrams assuming the angle A and load Z can change:
Y-hand & deviation images
physics
$endgroup$
What are the formulas for calculating the weight on X and Y in these diagrams assuming the angle A and load Z can change:
Y-hand & deviation images
physics
physics
edited Jan 12 at 18:47
Larry
2,39131129
2,39131129
asked Jan 12 at 17:20
cppVulkancppVulkan
1
1
closed as off-topic by John Douma, amWhy, Paul Frost, pre-kidney, John Bentin Jan 13 at 7:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Douma, amWhy, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by John Douma, amWhy, Paul Frost, pre-kidney, John Bentin Jan 13 at 7:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Douma, amWhy, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What have you tried?
$endgroup$
– Larry
Jan 12 at 17:32
$begingroup$
Nothing, I don't know where to start. I just want a couple of formulas so I can transfer them into c++
$endgroup$
– cppVulkan
Jan 12 at 18:15
add a comment |
1
$begingroup$
What have you tried?
$endgroup$
– Larry
Jan 12 at 17:32
$begingroup$
Nothing, I don't know where to start. I just want a couple of formulas so I can transfer them into c++
$endgroup$
– cppVulkan
Jan 12 at 18:15
1
1
$begingroup$
What have you tried?
$endgroup$
– Larry
Jan 12 at 17:32
$begingroup$
What have you tried?
$endgroup$
– Larry
Jan 12 at 17:32
$begingroup$
Nothing, I don't know where to start. I just want a couple of formulas so I can transfer them into c++
$endgroup$
– cppVulkan
Jan 12 at 18:15
$begingroup$
Nothing, I don't know where to start. I just want a couple of formulas so I can transfer them into c++
$endgroup$
– cppVulkan
Jan 12 at 18:15
add a comment |
1 Answer
1
active
oldest
votes
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You really should show attempts at problem-solving in your question but I will try to guide you with the kind of thinking you can do to help yourself with other problems. Think of the case where X and Y are the same point and angle-A is zero; then the load on each would be half of Z. Then think of X and Y forming a straight line with no sag; the load would be infinite. In between, the load varies inversely with sine. For the Y-hang, the angle(s) that X and Y have [downward] is $(90-frac{A}{2})$ and they each share half the load so I believe the load would be
$$Z=frac{1}{2}*frac{L}{sin(90-frac{A}{2})}$$
For the variation, the load is just going over a pully so X would feel the full force (L).
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You really should show attempts at problem-solving in your question but I will try to guide you with the kind of thinking you can do to help yourself with other problems. Think of the case where X and Y are the same point and angle-A is zero; then the load on each would be half of Z. Then think of X and Y forming a straight line with no sag; the load would be infinite. In between, the load varies inversely with sine. For the Y-hang, the angle(s) that X and Y have [downward] is $(90-frac{A}{2})$ and they each share half the load so I believe the load would be
$$Z=frac{1}{2}*frac{L}{sin(90-frac{A}{2})}$$
For the variation, the load is just going over a pully so X would feel the full force (L).
$endgroup$
add a comment |
$begingroup$
You really should show attempts at problem-solving in your question but I will try to guide you with the kind of thinking you can do to help yourself with other problems. Think of the case where X and Y are the same point and angle-A is zero; then the load on each would be half of Z. Then think of X and Y forming a straight line with no sag; the load would be infinite. In between, the load varies inversely with sine. For the Y-hang, the angle(s) that X and Y have [downward] is $(90-frac{A}{2})$ and they each share half the load so I believe the load would be
$$Z=frac{1}{2}*frac{L}{sin(90-frac{A}{2})}$$
For the variation, the load is just going over a pully so X would feel the full force (L).
$endgroup$
add a comment |
$begingroup$
You really should show attempts at problem-solving in your question but I will try to guide you with the kind of thinking you can do to help yourself with other problems. Think of the case where X and Y are the same point and angle-A is zero; then the load on each would be half of Z. Then think of X and Y forming a straight line with no sag; the load would be infinite. In between, the load varies inversely with sine. For the Y-hang, the angle(s) that X and Y have [downward] is $(90-frac{A}{2})$ and they each share half the load so I believe the load would be
$$Z=frac{1}{2}*frac{L}{sin(90-frac{A}{2})}$$
For the variation, the load is just going over a pully so X would feel the full force (L).
$endgroup$
You really should show attempts at problem-solving in your question but I will try to guide you with the kind of thinking you can do to help yourself with other problems. Think of the case where X and Y are the same point and angle-A is zero; then the load on each would be half of Z. Then think of X and Y forming a straight line with no sag; the load would be infinite. In between, the load varies inversely with sine. For the Y-hang, the angle(s) that X and Y have [downward] is $(90-frac{A}{2})$ and they each share half the load so I believe the load would be
$$Z=frac{1}{2}*frac{L}{sin(90-frac{A}{2})}$$
For the variation, the load is just going over a pully so X would feel the full force (L).
edited Jan 12 at 19:27
answered Jan 12 at 19:14
poetasispoetasis
419117
419117
add a comment |
add a comment |
1
$begingroup$
What have you tried?
$endgroup$
– Larry
Jan 12 at 17:32
$begingroup$
Nothing, I don't know where to start. I just want a couple of formulas so I can transfer them into c++
$endgroup$
– cppVulkan
Jan 12 at 18:15