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    How do I go about solving this question? Can anyone explain me? I have been up late trying solve it but no result so far.
    I attached the question.
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    (Original post by thebrahmabull)
    How do I go about solving this question? Can anyone explain me? I have been up late trying solve it but no result so far.
    I attached the question.
    Do you want to show us your working? You'll want to label the tension in the chains as T first of all, do some resolving and take moments.
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    Show us your diagram, so we can see if you've neglected/misplaced any forces


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    (Original post by Zacken)
    Do you want to show us your working? You'll want to label the tension in the chains as T first of all, do some resolving and take moments.

    (Original post by drandy76)
    Show us your diagram, so we can see if you've neglected/misplaced any forces


    Posted from TSR Mobile
    Thanks for looking! I wasn't sure where to start and my working may look completely rubbish. I have some questions :
    1.Is the force from the hinge directed along the draw bridge?
    2.Is the force from the hinge a kind of reaction force?
    3.in a 3 dimensional situation should forces directed at 2 directions count as perpendicular to an axis?
    4. Is there a way to solve the problem without resolving the force from the hinge into 2 components?
    If anyone could show a solution it would be great since I feel like a noob. Thanks!
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    1.Is the force from the hinge directed along the draw bridge?

    Supposing it was like that, should we prove it? I mean we should take into consideration that all the rest forces have components vertical to the draw bridge's plane

    3.in a 3-dimensional situation should forces directed at 2 directions count as perpendicular to an axis?
    Let as set our three-dimensional cartesian coordinate system so as its z-axis is along the direction which is defined by AD. Consider the directions of the forces: Tension, weight. Has any of them a component along z axis?

    4. Is there a way to solve the problem without resolving the force from the hinge into 2 components?Maybe, but ...not resolving the force from the hinge into two components does not make it simpler.


    (Original post by thebrahmabull)
    Thanks for looking! I wasn't sure where to start and my working may look completely rubbish. I have some questions :1.Is the force from the hinge directed along the draw bridge?2.Is the force from the hinge a kind of reaction force?3.in a 3 dimensional situation should forces directed at 2 directions count as perpendicular to an axis?4. Is there a way to solve the problem without resolving the force from the hinge into 2 components?If anyone could show a solution it would be great since I feel like a noob. Thanks!
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    (Original post by depymak)
    1.Is the force from the hinge directed along the draw bridge?

    Supposing it was like that, should we prove it? I mean we should take into consideration that all the rest forces have components vertical to the draw bridge's plane

    3.in a 3-dimensional situation should forces directed at 2 directions count as perpendicular to an axis?
    Let as set our three-dimensional cartesian coordinate system so as its z-axis is along the direction which is defined by AD. Consider the directions of the forces: Tension, weight. Has any of them a component along z axis?

    4. Is there a way to solve the problem without resolving the force from the hinge into 2 components?Maybe, but ...not resolving the force from the hinge into two components does not make it simpler.
    Thank you! How can I prove the direction of the force from the hinge?
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    Take moments about AD axis, as ZacKen proposed. You don't need the force from the hinge at this stage. Have you determined the magnitude of T?
    For the next step ,I draw axes according to my opinion. Weight has the direction of Y axis. Let name the unknown force R. I think that R hasn't projection onto z-axis. (Rz=0). Does it make sense to you?
    Name:  AXIS.JPG
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    (Original post by Zacken)
    Do you want to show us your working? You'll want to label the tension in the chains as T first of all, do some resolving and take moments.
    (Original post by drandy76)
    Show us your diagram, so we can see if you've neglected/misplaced any forcesPosted from TSR Mobile
    (Original post by depymak)
    Take moments about AD axis, asZacKen proposed. You don't need the force from the hinge at this stage. Haveyou determined the magnitude of T?For the next step ,I draw axes according to my opinion.Weight has the direction of Y axis. Let name the unknown force R. I think thatR hasn't projection onto z-axis. (Rz=0). Does it make sense to you?
    Thanks. I made an attempt without taking into account the exactdirection of force from the hinge.First I found the angle theta using trigonometry as shown.Having done that, I resolved tension into 2 components. Then I took momentsabout AD to calculate tension. Then assuming tension, weight and force from thehinge (r) are the only forces acting, I used the cosine rule to find r whichgives me 8870N. But the correct answer is 7470 at 48 degree to horizontal. Any suggestions?
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    Good job! One objection, though..
    The way you resolved tension ....I think this way you miscalculate distances when you try to define moments. Moreover, the vector triangle should be a closing figure (the same way you did it when you found x....which by the way answered your previous question: you can indeed calculate the unknown force without resolving it)

    (Original post by thebrahmabull)
    Thanks. I made an attempt without taking into account the exactdirection of force from the hinge.First I found the angle theta using trigonometry as shown.Having done that, I resolved tension into 2 components. Then I took momentsabout AD to calculate tension. Then assuming tension, weight and force from thehinge (r) are the only forces acting, I used the cosine rule to find r whichgives me 8870N. But the correct answer is 7470 at 48 degree to horizontal. Any suggestions?
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    (Original post by depymak)
    Good job! One objection, though..
    The way you resolved tension ....I think this way you miscalculate distances when you try to define moments. Moreover, the vector triangle should be a closing figure (the same way you did it when you found x....which by the way answered your previous question: you can indeed calculate the unknown force without resolving it)
    Thanks . I didnot quite get what you meant :?
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    sorry, I hurried.
    You can also resolve T this way. What do you think?
    Name:  m2 question.jpg
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    all the rest work of yours is fine
    If I were you, I would also define the direction angle of the unknown force. Why not using the law of sines?I liked the triangle of forces..very useful.
    (Original post by thebrahmabull)
    Thanks . I didnot quite get what you meant :?
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    (Original post by depymak)
    sorry, I hurried.
    You can also resolve T this way. What do you think?
    Name:  m2 question.jpg
Views: 142
Size:  178.3 KB
    all the rest work of yours is fine
    If I were you, I would also define the direction angle of the unknown force. Why not using the law of sines?I liked the triangle of forces..very useful.
    Wow! Thanks so much so the fact that I took 7m. as the distance of 2tsintheta from AD messed it up. Is there any rule I should follow to avoid mistakes like these? Like something coming in my mind is: taking the closest possible distance of the force from the pivot.

    Well now that all 3 forces are in place the triangle of forces can be used once again to define the direction
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    Did it work? I am very happy.
    What came in your mind is very close to what I think. You said the closest/shortest (perpendicular) distance from the pivot to the line of the action of the force, which sounds well. I like it.
    I found this demonstration
    http://demonstrations.wolfram.com/Mo...ceAboutAPoint/
    In case you are going to use it, just find tension's moment by yourself and then check it.
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    (Original post by depymak)
    Did it work? I am very happy.
    What came in your mind is very close to what I think. You said the closest/shortest (perpendicular) distance from the pivot to the line of the action of the force, which sounds well. I like it.
    I found this demonstration
    http://demonstrations.wolfram.com/Mo...ceAboutAPoint/
    In case you are going to use it, just find tension's moment by yourself and then check it.
    yep worked I will check out the link. thanks for all the help!
 
 
 
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