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M1 aqa forces

I just can't get to grips with this question.

two light inextensible strings are attached to a particle of mass m kg. each string passes over a a fixed smooth pulley. The other end of one string is attached to a particle of mass 4 kg. the other end of the second string is attached to a particle of mass 3 kg. the diagram shows an m shape- sorry can't draw it. The 4 kg side of the mass has an angle of 43 deg to the vertical and the 3 side of the mass has an angle of theta to the vertical.

A) calculate the tension in the string - done
b) show that theta is 65.4 degrees. - no idea. Tried all sorts and haven't got a clue.
c) find m - even with the answer for b) I can't sort it.


help gratefully awaited.
On b, if you know what the net resultant force is parallel to the plane then you are able to figure out the angle (if you don't then see if the question tells you it's in equilibrium. If you know the force parallel to the plane at one end and know the resultant force, it's a simple case of subtracting when force from the other and solving for the angle

So
4gCos43−3gCosθ=Net  ResultantForce4gCos43 - 3gCos\theta = Net\ \ Resultant Force

If you can take a picture of the full question that might be helpful :smile: Not 100% on what I just put but hopefully it puts you on the right tracks anyway :smile:
I don't know. There isn't a net resultant force is there as it's in equilibrium. I thought that perhaps t1 sin 43 = t2 sin theta and I have the values for the tensions but it doesn't come out to 65.

I can't send a picture I'm using an iPad and have no idea about that either. Hopeless aren't i. But on the web there is that exact question in a book called revise for mechanics 1. It's question 29 in the third chapter.

i know tar isn't supposed to give answers but having fiddled about with every conceivable set of numbers I cannot find the solution. I have done the previous 28 questions. Struggled but done.
Original post by maggiehodgson
I just can't get to grips with this question.

two light inextensible strings are attached to a particle of mass m kg. each string passes over a a fixed smooth pulley. The other end of one string is attached to a particle of mass 4 kg. the other end of the second string is attached to a particle of mass 3 kg. the diagram shows an m shape- sorry can't draw it. The 4 kg side of the mass has an angle of 43 deg to the vertical and the 3 side of the mass has an angle of theta to the vertical.

A) calculate the tension in the string - done
b) show that theta is 65.4 degrees. - no idea. Tried all sorts and haven't got a clue.
c) find m - even with the answer for b) I can't sort it.


help gratefully awaited.


Tension in string attached to 3g mass will be 3g N and that in other string will be 4g N. Resolve horizontally for the particle of mass m then we have
3sin⁡θ=4sin⁡43o 3\sin\theta=4\sin 43^o
Resolve vertically to find m.
Hi and thanks. That does give the correct answer. It has taken me a while to work out why that is the case but I get it now.

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