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# M1 problem watch

1. A small sign of weight 6N is suspended by two chains AC and BC, as shown in the diagram.

A and B are attached to fixed points which are 60cm apart and at the same horizontal level. The lengths of AC and BC are 40cm and 30cm.

Calculate the tension in the chains.It's probably not very hard, but if the vertical component of the two triangles of hypotenuses 30 cm and 40 cm is in N - 6 N - surely you can't calculate the force? I am missing something... thanks!
2. (Original post by Jmun)
It's probably not very hard, but if the vertical component of the two triangles of hypotenuses 30 cm and 40 cm is in N - 6 N - surely you can't calculate the force? I am missing something... thanks!
I'm sorry I don't understand this part (in bold), what do you mean by "is in N"?

Although I do know one thing, this question is doable.
3. Tension acts away from the mass.
Split AC and BC in horizontal and vertical components.
the two downward will have a force downwards summed to 6N.
use F=ma, where a=9.8ms^-2.
or W=mg
Hope some of that might help.
4. (Original post by gagafacea1)
I'm sorry I don't understand this part (in bold), what do you mean by "is in N"?

Although I do know one thing, this question is doable.
I think it means the vertical component is measured in newtons.
5. (Original post by laurenejones)
I think it means the vertical component is measured in newtons.
I did think about that, but it still doesn't make sense to me.
6. (Original post by gagafacea1)
I did think about that, but it still doesn't make sense to me.
yeah, the problem's that the measurements given are in cm, and it would have to be converted to newtons in some way. the two downward components would add to 6N, and the length should not have an effect. can work out the mass of the object if that would be helpful: (6/9.8)g
7. (Original post by laurenejones)
yeah, the problem's that the measurements given are in cm, and it would have to be converted to newtons in some way. the two downward components would add to 6N, and the length should not have an effect. can work out the mass of the object if that would be helpful: (6/9.8)g
Oh I'm sorry, I meant that sentence doesn't make sense to me. I actually know how to do this question. But we're not supposed to provide full solutions. And the OP hasn't replied at all, so I'm refraining from posting anything until they do.
8. Thanks guys, I'll try it out and let you know if I still struggle...
Thanks lots!
9. (Original post by laurenejones)
yeah, the problem's that the measurements given are in cm, and it would have to be converted to newtons in some way. the two downward components would add to 6N, and the length should not have an effect. can work out the mass of the object if that would be helpful: (6/9.8)g

(Original post by Jmun)
Thanks guys, I'll try it out and let you know if I still struggle...
Thanks lots!
The distance measurements aren't really an issue here. You need some angles so that you can resolve forces, and the cosine rule will give you one angle in the triangle from which you can derive any of the other angles from the sine rule if necessary).

Once you know some angles you can write equations for the horizontal and vertical components of the tensions.
10. (Original post by Jmun)
A small sign of weight 6N is suspended by two chains AC and BC, as shown in the diagram.

A and B are attached to fixed points which are 60cm apart and at the same horizontal level. The lengths of AC and BC are 40cm and 30cm.

Calculate the tension in the chains.It's probably not very hard, but if the vertical component of the two triangles of hypotenuses 30 cm and 40 cm is in N - 6 N - surely you can't calculate the force? I am missing something... thanks!
Call the tensions T1 and T2 and then resolve vertically and horizontally as the system is in equilibrium. [not F=ma .... ]

The distances are there to help you resolve - draw a line vertically through C.

If you post something then I'll check it
11. (Original post by davros)
The distance measurements aren't really an issue here. You need some angles so that you can resolve forces, and the cosine rule will give you one angle in the triangle from which you can derive any of the other angles from the sine rule if necessary).

Once you know some angles you can write equations for the horizontal and vertical components of the tensions.
(Original post by Muttley79)
Call the tensions T1 and T2 and then resolve vertically and horizontally as the system is in equilibrium. [not F=ma .... ]

The distances are there to help you resolve - draw a line vertically through C.

If you post something then I'll check it
I worked out the angles:
(let the point where the vertical line through C cuts AB be O)
^ACB = 117.2;
^CAO = 26.4;
^BCO = 36.4.
Now im stuck again....
If i try use the 6N as given to find the tension like this for example:
6 / sin 36.4 = 10.11 which is wrong..
12. (Original post by Jmun)
I worked out the angles:
(let the point where the vertical line through C cuts AB be O)
^ACB = 117.2;
^CAO = 26.4;
^BCO = 36.4.
Now im stuck again....
If i try use the 6N as given to find the tension like this for example:
6 / sin 36.4 = 10.11 which is wrong..
Resolve forces vertically - there are two tensions ...

component of T1 + component of T2 = 6

Then horizontally

component of T1 = component of T2

Post what you get at each step
13. Thanks so much to everyone!!! I figured it out!!!
I put T1, T2 an 6N in a closed force diagram, worked out the angles using cosine rule and sin rule aa well as T1 and T2 using the sine rule! I was over complicating it.
T1 = 5.4N T2 = 6N
Thanks so much everyone!!!

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