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# Stuck on M2 moments watch

1. Question attached. For a), I called tension on string BC Ta and tension in D and the mass Tb.

Resolve upwards:
Tasinα+Tb=600

Moment around a):
400(2)+200(3)=Tasinα(4)+Tb(3)
Which leads to Ta=400root5 when solving simultaneously

Where have I gone wrong?

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2. (Original post by bobbricks)
Question attached. For a), I called tension on string BC Ta and tension in D and the mass Tb.

Resolve upwards:
Tasinα+Tb=600

Moment around a):
400(2)+200(3)=Tasinα(4)+Tb(3)
Which leads to Ta=400root5 when solving simultaneously

Where have I gone wrong?

Posted from TSR Mobile
I get 400(2)+200(3) = 4Tsina by taking moments about the hinge

then resolving

Verically
Y+Tsina = 600

Horizontally
X=Tcosa

where X and Y are the components of reaction at the hinge
3. (Original post by TeeEm)
I get 400(1.5)+200(2) = 3Tsina by taking moments about the hinge
I think this should be 400(2) + 200(3) = 4Tsina.
4. (Original post by SherlockHolmes)
I think this should be 400(2) + 200(3) = 4Tsina.
possibly
i might have misread distances

teaching at the moment and I have no glasses!...
5. (Original post by TeeEm)
I get 400(2)+200(3) = 4Tsina by taking moments about the hinge

then resolving

Verically
Y+Tsina = 600

Horizontally
X=Tcosa

where X and Y are the components of reaction at the hinge
Isn't the reaction at the hinge horizontal so no vertical component?

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6. (Original post by bobbricks)
Isn't the reaction at the hinge horizontal so no vertical component?

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Definitely not!
Consider what will happen if you take moments about the other end of the rod ...
What balances the two weights then?
7. (Original post by TeeEm)
Definitely not!
Consider what will happen if you take moments about the other end of the rod ...
What balances the two weights then?
The vertical component of the string BC (Tsinalpha)?

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8. (Original post by bobbricks)
The vertical component of the string BC (Tsinalpha)?

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if you take moments about B, the tension has no moment!
So there is vertical reaction (UP) to balance the 2 weights.

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