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    Hi,
    This question attached is really confusing me.
    Why on earth is the vertical force exerted by the rod 60N??
    It is really confusing me?
    And the tension question too??
    I've tried it but not got to either conclusions correctly!
    Thanks.

    Update I have got part b but part a is the one I do not understand.
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    (Original post by nwmyname)
    Hi,
    This question attached is really confusing me.
    Why on earth is the vertical force exerted by the rod 60N??
    It is really confusing me?
    And the tension question too??
    I've tried it but not got to either conclusions correctly!
    Thanks.
    Take moments about the point where the rod meets the wall, so that the unknown reaction force isn't required.

    Call the length of the rod L.

    The weight of the sign acts at a distance L/2, so its moment is L/2 * 120.

    The vertical component of the tension has a moment equal to L * Tsin(30), where Tsin(30) is the the vertical force. To simplify this, you could call this F, but you will need Tsin(30) for part B.

    Moment clockwise = moment anticlockwise.
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    Have you covered moments?
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    (Original post by ombtom)
    Take moments about the point where the rod meets the wall, so that the unknown reaction force isn't required.

    Call the length of the rod L.

    The weight of the sign acts at a distance L/2, so its moment is L/2 * 120.

    The vertical component of the tension has a moment equal to L * Tsin(30), where Tsin(30) is the the vertical force. To simplify this, you could call this F, but you will need Tsin(30) for part B.

    Moment clockwise = moment anticlockwise.
    So to get this right, the clockwise moment is given by L/2 * 120, ok. How does this help me find the other moment?
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    (Original post by Thisguy11)
    Have you covered moments?
    yes, I understand that. But how do I calculate the anticlockwise moment?
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    (Original post by nwmyname)
    yes, I understand that. But how do I calculate the anticlockwise moment?
    I dont think you need that, isnt total moment =0?, there is no roation?
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    (Original post by nwmyname)
    So to get this right, the clockwise moment is given by L/2 * 120, ok. How does this help me find the other moment?
    The other force on the diagram is the tension. Resolve it into vertical and horizontal components using sin and cos; only the vertical component will have a moment about the chosen point.
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    (Original post by ombtom)
    The other force on the diagram is the tension. Resolve it into vertical and horizontal components using sin and cos; only the vertical component will have a moment about the chosen point.
    I think I've got it. Something like this?
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    (Original post by nwmyname)
    I think I've got it. Something like this?
    Yes; the red arrow going up is the vertical component of the tension – Tsin(30).
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    (Original post by ombtom)
    Yes; the red arrow going up is the vertical component of the tension – Tsin(30).
    Great! So the two forces on the diagrams must be equal?

    The vertical tension must equal the weight?

    If this is right, then I have solved it! Hopefully!
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    (Original post by ombtom)
    Yes; the red arrow going up is the vertical component of the tension – Tsin(30).
    Oh good lord. This question as well is giving me the stress. I'd be grateful if you could give me a clue or a starting point to solving this!
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    (Original post by nwmyname)
    Great! So the two forces on the diagrams must be equal?

    The vertical tension must equal the weight?

    If this is right, then I have solved it! Hopefully!
    Unfortunately there are also reaction forces against the wall. You have to use your moments equation to find T, as I said above... 1/2 * 120 = T sin(30).
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    (Original post by ombtom)
    Unfortunately there are also reaction forces against the wall. You have to use your moments equation to find T, as I said above... 1/2 * 120 = T sin(30).
    T would still be 120N.
    Oh I already calculated that.
    It was only a) that I was struggling at.

    OHHHH.
    So you calculate the tension first, then use that to find the vertical component???
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    (Original post by nwmyname)
    Oh good lord. This question as well is giving me the stress. I'd be grateful if you could give me a clue or a starting point to solving this!
    There are three forces involved. For equilibrium, they must all pass through the same point.

    The three forces are the tension, the weight, and the reaction at the point where the rod meets the wall.

    Are you okay with the first part now?
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    (Original post by nwmyname)
    So you calculate the tension first, then use that to find the vertical component???
    Other way around!
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    (Original post by ombtom)
    Other way around!
    My bad! Apologies.

    L/2 * 120 = F * L
    60 L = F * L
    F = 60.

    Correct now?

    Like you mentioned, consider Tsin30 = F.
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    (Original post by nwmyname)
    My bad! Apologies.

    L/2 * 120 = F * L
    60 L = F * L
    F = 60.

    Correct now?

    Like you mentioned, consider Tsin30 = F.
    Perfect.
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    (Original post by ombtom)
    perfect. :e
    yaaaaaay!
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    (Original post by ombtom)
    There are three forces involved. For equilibrium, they must all pass through the same point.

    The three forces are the tension, the weight, and the reaction at the point where the rod meets the wall.

    Are you okay with the first part now?
    That's really got me stuck.
    The weight is down, the tension is at the angle, but where does the third one go? I am so stuck it's unbelieveable.
    I'm really urging myself not to look at the mark scheme!
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    (Original post by nwmyname)
    That's really got me stuck.
    The weight is down, the tension is at the angle, but where does the third one go? I am so stuck it's unbelieveable.
    I'm really urging myself not to look at the mark scheme!
    Extend the arrow for weight until it passes through the arrow for tension.

    You know that the reaction force must pass through the point where the rod meets the wall.

    + That's a good attitude to have.
 
 
 
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