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Need help with alevel maths mechanics question

ive spent hours on this question - i do have an answer but it looks soo wrong.
What am i doing wrong? ive tried resolving the forces vertically and horizontally and resolving the moments about A (on my diagram).
can someone please show me how to do it

edit: i got the question off my friend and i noticed it doesnt say anything about being in equilibrum but i assumed it is because he prob skipped it out while reading it to me
(edited 2 years ago)
maths quest.pdf995.6KB
Reply 1
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mqb2766
21
Original post by Ayo.f1
ive spent hours on this question - i do have an answer but it looks soo wrong.
What am i doing wrong? ive tried resolving the forces vertically and horizontally and resolving the moments about A (on my diagram).
can someone please show me how to do it

edit: i got the question off my friend and i noticed it doesnt say anything about being in equilibrum but i assumed it is because he prob skipped it out while reading it to me

How do you take moments about A?
Original post by Ayo.f1
ive spent hours on this question - i do have an answer but it looks soo wrong.
What am i doing wrong? ive tried resolving the forces vertically and horizontally and resolving the moments about A (on my diagram).
can someone please show me how to do it

edit: i got the question off my friend and i noticed it doesnt say anything about being in equilibrum but i assumed it is because he prob skipped it out while reading it to me

Solve simultaneously for reaction forces from your first two equations.

Also where is the 6 coming from in your moment equation?
Reply 3
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Ayo.f1
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Original post by mqb2766
How do you take moments about A?

thats the only method i can think of and i can see that there are turning forces at that point so it should it on the right line i thought. I couldve taken moments about B but thatll be harder since i dont have an angle there
Reply 4
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mqb2766
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Original post by Ayo.f1
thats the only method i can think of and i can see that there are turning forces at that point so it should it on the right line i thought. I couldve taken moments about B but thatll be harder since i dont have an angle there

I think RDKGames gave a less subtle hint. Where does the 6 come from?

Tbh, its a reasonably common exam question and there are quite a few very similar mark schemes from past papers you could look at.
(edited 2 years ago)
Reply 5
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Ayo.f1
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Original post by RDKGames
Solve simultaneously for reaction forces from your first two equations.

Also where is the 6 coming from in your moment equation?

whoops that should say 3m - the length of the rod from the questions, i doubled it for some reason. ive corrected it but that makes no difference to my working out.
i cant see how to solve simultaneously because the two equations have different variables

edit: the horizontal equations is Rb - Fa = 0 and the vertical is Fb + Ra - 25g = 0 so nothing in common
ive done all the sensible substitutions in the moments equation i could see
anything else i cant see?
(edited 2 years ago)
Original post by Ayo.f1
whoops that should say 3m - the length of the rod from the questions, i doubled it for some reason. ive corrected it but that makes no difference to my working out.
i cant see how to solve simultaneously because the two equations have different variables

Yeah but friction forces you know them in terms of the very same reaction forces.

This is ultimately a system of two equations in two unknowns.
(edited 2 years ago)
Reply 7
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Ayo.f1
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Original post by RDKGames
Yeah but friction forces you know them in terms of the very same reaction forces.

This is ultimately a system of two equations in two unknowns.

thanks for your response i definetly agree but i cant seem to figure it out. ive been trying literally all day
can you see anything in my working that might be wrong?

edit: or maybe because there is 5 unknowns in there maybe the answer is in terms of of one of the other unknowns like my answer?
(edited 2 years ago)
Original post by Ayo.f1
thanks for your response i definetly agree but i cant seem to figure it out. ive been trying literally all day
can you see anything in my working that might be wrong?

Method is good, just the 6 you had was wrong.

By finding the reaction forces you simplify tan of the angle to a single constant.

So that's what you need to do; find the reaction forces of this system then sub in the relevant one at the end in your answer.
(edited 2 years ago)
Reply 9
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Ayo.f1
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Original post by RDKGames
; find the reaction forces of this system then sub in the relevant one at the end in your answer.

im not sure how to find the reaction forces. would that maybe be the component of the weight of the rod?
Original post by Ayo.f1
im not sure how to find the reaction forces. would that maybe be the component of the weight of the rod?

Your system:

{RB−FA=0FB+RA−25g=0\begin{cases} R_B - F_A = 0 \\ F_B + R_A - 25g = 0 \end{cases}

You know: FA=45RAF_A = \dfrac{4}{5}R_A and FB=35RBF_B = \dfrac{3}{5}R_B

The system becomes:

{RB−45RA=035RB+RA−25g=0\begin{cases} R_B - \dfrac{4}{5}R_A = 0 \\ \dfrac{3}{5}R_B + R_A - 25g = 0 \end{cases}


Solve simultaneously just like at GCSE level ....
(edited 2 years ago)
Reply 11
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Ayo.f1
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Original post by RDKGames
Your system:

{RB−FA=0FB+RA−25g=0\begin{cases} R_B - F_A = 0 \\ F_B + R_A - 25g = 0 \end{cases}

You know: FA=45RAF_A = \dfrac{4}{5}R_A and FB=35RBF_B = \dfrac{3}{5}R_B

The system becomes:

{RB−45RA=035RB+RA−25g=0\begin{cases} R_B - \dfrac{4}{5}R_A = 0 \\ \dfrac{3}{5}R_B + R_A - 25g = 0 \end{cases}


Solve simultaneously just like at GCSE level ....

OMG!
i am sooo ashamed of myself ! :mad:
Thank you sooo much for your help!! for some reason i totally didnt see that
(edited 2 years ago)
Original post by Ayo.f1
OMG!
i am sooo ashamed of myself ! :mad:
Thank you sooo much!! for some reason i didnt see that

Long day, eh?

On your upload you effectively have five unknowns BUT you also have five equations in involving these unknowns. So there is no degree of freedom here. If there was, then yes tan would have to be expressed in terms of another unknown(s).
(edited 2 years ago)
Reply 13
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Ayo.f1
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Original post by RDKGames
Long day, eh?

On your upload you effectively have five unknowns BUT you also have five equations in involving these unknowns. So there is no degree of freedom here. If there was, then yes tan would have to be expressed in terms of another unknown(s).

oh yes its been a what feels like wasted day now that see how straightforward that part is. i did that part since morning and just couldnt see what i had done wrong thinking my diagram is wrong, forces are wrong my brain just went all foggy i worked the samething out so many times
i seriously need to open my eyes more
(edited 2 years ago)

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