# Hardest Mechanics Question Ever - Can you solve?

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#1
Hi, so I (and 10 others) have been stuck on a Mechanics Question shown below - part cii, in particular. I was wondering whether anyone on TSR could solve it. Thank you to everyone in advance - I have spent hours on it to no avail! :/

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1 month ago
#2
is this Y2 maths or uni?
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1 month ago
#3
Assuming you've done the other parts correctly? What do you get for the basic forces equation with theta < alpha? You're really just bounding the terms in b)
Last edited by mqb2766; 1 month ago
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#4
(Original post by Timo werner)
is this Y2 maths or uni?
This is Y2 Maths
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#5
(Original post by mqb2766)
Assuming you've done the other parts correctly? What do you get for the basic forces equation with theta < alpha? You're really just bounding the terms in b)
The other parts were pretty easy.

What do you mean by 'bounding the terms in b)'?

I got:

Mbg - T = MbA (A = acceleration)

T - (mu*Magcos(theta) + Magsin(theta)) = MaA

?

EDIT: Left out mu accidentally.
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1 month ago
#6
(Original post by TheWilkerWay)
The other parts were pretty easy.

What do you mean by 'bounding the terms in b)'?

I got:

Mbg - T = MbA (A = acceleration)

T - (mu*Magcos(theta) + Magsin(theta)) = MaA

?

EDIT: Left out mu accidentally.
So eliminate T and get A = ...
Last edited by mqb2766; 1 month ago
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#7
(Original post by mqb2766)
So eliminate T and get A = ...
I got A = (Mbg - (mu*Ma*g*cos(theta)) - (Ma*g*sin(theta)) / (Ma + Mb)
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1 month ago
#8
(Original post by TheWilkerWay)
I got A = (Mbg - (mu*Ma*g*cos(theta)) - (Ma*g*sin(theta)) / (Ma + Mb)
put the expression for mu and that's the first part done.
then factor out g and look at the desired expression, what bounding can you do?
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#9
(Original post by mqb2766)
put the expression for mu and that's the first part done.
then factor out g and look at the desired expression, what bounding can you do?
Not quite sure about the bounding - could you perhaps give a hint?

I have attached an image of what I have so far - am I correct?
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1 month ago
#10
For the 2nd term in the numerator
* Its negative
* Cos (theta) > cos(alpha)
You want to make the expression simple and a lower bound for the actual acceleration, so what could you replace it by?

Is the tan term positive?
Last edited by mqb2766; 1 month ago
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#11
(Original post by mqb2766)
For the 2nd term in the numerator
* Its negative
* Cos (theta) > cos(alpha)
You want to make the expression simple and a lower bound for the actual acceleration, so what could you replace it by?

Is the tan term positive?
Tan term would be negative. as tanx and cos x in 0 < x < 90 would always be positive and there is a negative infront of the Ma?

I am still so confused, sorry.

The only thing that I can think of is that all the negative terms will always be less than zero?
Last edited by TheWilkerWay; 1 month ago
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1 month ago
#12
(Original post by TheWilkerWay)
Tan term would be negative.

I am still so confused, sorry.
Cos is decreasing on 0..90 (first part of question. Sin is increasing.
Tan term is pos as it's double negative?
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#13
(Original post by mqb2766)
Cos is decreasing on 0..90 (first part of question. Sin is increasing.
Tan term is pos as it's double negative?
Yeah I was stupid on both. I forgot the double negative.

I am still not sure how to solve it via bounding. The only thing I can think of is somehow comparing the two terms with Mb with each other and to do the same for the two terms with Ma.
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1 month ago
#14
(Original post by TheWilkerWay)
Yeah I was stupid on both. I forgot the double negative.

I am still not sure how to solve it via bounding. The only thing I can think of is somehow comparing the two terms with Mb with each other and to do the same for the two terms with Ma.
Take the tan(alpha) term. What is the minimum it can be?
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#15
(Original post by mqb2766)
Take the tan(alpha) term. What is the minimum it can be?
Well tan alpha would tend to zero if alpha and theta both tend to zero?
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1 month ago
#16
(Original post by TheWilkerWay)
Well tan alpha would tend to zero if alpha and theta both tend to zero?
The game is to fix alpha (the answer includes it) but remove theta and masses from the expression by boundin g. Can you upload the expression you're working withnow?
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#17
(Original post by mqb2766)
The game is to fix alpha (the answer includes it) but remove theta and masses from the expression by boundin g. Can you upload the expression you're working withnow?
I might just be stupid as my expression is the same as before. I don't really understand how to eliminate parts of the expression by using limits or inequalities.
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1 month ago
#18
(Original post by TheWilkerWay)
I might just be stupid as my expression is the same as before. I don't really understand how to eliminate parts of the expression by using limits or inequalities.
Tan sign changed?
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1 month ago
#19
You want to bound things. Have a go at getting (bounding) the numerator as something like
mA(1 - tan(alpha)) + mB(1 - tan(alpha))
The rest then follows. So group the two mA terms together and the two MB terms.
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1 month ago
#20
That is an old version of the question which I had corrected a few days after posting it.

The corrected version of the question is here: https://www.thestudentroom.co.uk/sho...&postcount=225

The game remains the same though, you bound variables and expressions in some way to obtain the lower bound for .
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