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Physics - Moments Question

Bit stumped on this question.

Fig 2 (below) shows a vase placed on a uniform shelf that is supported by a steel wire (orange line below). The ma ss of the vase is 0.65kg and the mass of the shelf is 2.0kg. The shelf is hinged at A. The steel wire is attached to the shelf 0.30m from A and is at an angle of 30 degrees to the shelf. The other end of the steel wire is attached to the wall.

Show, by taking moments about A, that the tension in the steel wire is about 50 N

So I thought I would take moments about A as the question asked and I did

A = (9,81*0.65*0.45) + (2*9.81*0.25) [0.25 because uniform beam so all mass acts here]

Then I thought I would get the force on A and then do A/sin(30). I got a value that was no where near 50N. Anyone have any ideas? Thanks.
Your working is correct. But let's not forget that the string is anchored to the shelf 0.3m away.
Therefore it is not Asin30\frac{A}{\sin 30}, it is the counterclockwise moment of A0.3sin30\frac{A}{0.3\sin 30}
Which will yield 51N :smile:
Original post by The-Spartan
Your working is correct. But let's not forget that the string is anchored to the shelf 0.3m away.
Therefore it is not Asin30\frac{A}{\sin 30}, it is the counterclockwise moment of A0.3sin30\frac{A}{0.3\sin 30}
Which will yield 51N :smile:


Thanks for the reply :smile:

I sort of understand why you did *0.3 but I don't know why you did it to the sin(30)?
I thought the force acting on A would just be trig and not involve the 0.3? I'm still a bit confused! Sorry if I sound stupid! I just suck at the mech part of phys :frown:
Original post by CrazyFool229
Thanks for the reply :smile:

I sort of understand why you did *0.3 but I don't know why you did it to the sin(30)?
I thought the force acting on A would just be trig and not involve the 0.3? I'm still a bit confused! Sorry if I sound stupid! I just suck at the mech part of phys :frown:


Here, ill draw a diagram for you just a sec, ill edit this in a min.
And no you dont sound stupid at all! its not easy stuff! :biggrin:
(edited 7 years ago)
Original post by The-Spartan
Here, ill draw a diagram for you just a sec, ill edit this in a min.
And no you dont sound stupid at all! its not easy stuff! :biggrin:


Wow thanks for that. Much appreciated :biggrin:
So just to clarify, is the clockwise moment A the dotted arrow going upwards then?
This looks like a question from one of the Physics specimen papers. Nice
Original post by CrazyFool229
Wow thanks for that. Much appreciated :biggrin:
So just to clarify, is the clockwise moment A the dotted arrow going upwards then?


No worries :smile:
The clockwise moment is the one you found yourself. You labelled this A so i did too, maybe you have anticlockwise and clockwise mixed up?

The anti-clockwise moment is the dotted line, acting upwards. This is T(0.3sin30)T(0.3\sin{30}).
Original post by TheTechnoGuy
This looks like a question from one of the Physics specimen papers. Nice


Yup, it is. Just got round to finishing all the questions but I had to go back to this one because I couldn't get anywhere near 50N! Turns out it was that one silly step with the *0.3 :frown:
Original post by The-Spartan
No worries :smile:
The clockwise moment is the one you found yourself. You labelled this A so i did too, maybe you have anticlockwise and clockwise mixed up?

The anti-clockwise moment is the dotted line, acting upwards. This is T(0.3sin30)T(0.3\sin{30}).


Yep, my bad again getting it mixed up haha. Thanks a lot for your help btw! :biggrin:
Original post by CrazyFool229
Yep, my bad again getting it mixed up haha. Thanks a lot for your help btw! :biggrin:


I do it all the time :wink:
And it's no problem! :biggrin: glad to be of help.
You need to work out the vertical force upwards from the rope A(7.78)=0.3*? So ?=25.9 Then just use trigonometry to find the force along the wire which is at 60 to the vertical
Reply 11
thank you!

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