The Student Room Group

M1 Reaction question

I understand how to do the question but I don't understand why the reaction at D (Q) isn't perpendicular to the pole. Can someone explain why?

Screen Shot 2016-05-22 at 13.11.03.png
Reply 1
Original post by ozmo19
I understand how to do the question but I don't understand why the reaction at D (Q) isn't perpendicular to the pole. Can someone explain why?

Screen Shot 2016-05-22 at 13.11.03.png


I'm uncertain as to why you think D (Q) is a reaction force? It's just a random force, not a reaction one. Hence there's no reason to believe it's perpendicular to the pole.
Reply 2
Original post by Zacken
I'm uncertain as to why you think D (Q) is a reaction force? It's just a random force, not a reaction one. Hence there's no reason to believe it's perpendicular to the pole.


Ahh, I misread the question when it said P was at a right angle, I assumed they meant Q was as well. Thanks.
I have another question. In 6(iii) when finding the integral of v to find an expression for s the mark scheme states that the constant cancels for the times given. I don't understand this as Ive never encountered it. Can you explain this?

Screen Shot 2016-05-22 at 13.23.25.png
(edited 7 years ago)
Original post by ozmo19
Ahh, I misread the question when it said P was at a right angle, I assumed the meant Q was as well. Thanks.
I have another question. In 6(iii) when finding the integral of v to find an expression for s the marks scheme states that the constant cancels for the times given. I don't understand this as Ifve never encountered it. Can you explain this?

Screen Shot 2016-05-22 at 13.23.25.png


It's always the same when you are evaluating between limits. Say you were doing [x^2 +7x +c] with bounds 4 and 0. You would get (4^2 +28 +c) -(0^2+0+c) and as you can see the c cancels
Reply 4
Original post by ozmo19
In 6(iii) when finding the integral of v to find an expression for s the marks scheme states that the constant cancels for the times given. I don't understand this as Ive never encountered it. Can you explain this?

Screen Shot 2016-05-22 at 13.23.25.png


If you've ever integrated between limits, you should be aware of this :/

If not ~ then you should find that the constant will cancel anyway.
Reply 5
Original post by ozmo19
Ahh, I misread the question when it said P was at a right angle, I assumed the meant Q was as well. Thanks.
I have another question. In 6(iii) when finding the integral of v to find an expression for s the marks scheme states that the constant cancels for the times given. I don't understand this as Ive never encountered it. Can you explain this?

Screen Shot 2016-05-22 at 13.23.25.png


Are you familiar with the idea that displacement is the area under the velocity-time graph?

Are you familiar with the idea that area under a graph f(x)f(x) between x=ax=a and x=bx=b is abf(x)dx\int_a^b f(x) \, \mathrm{d}x?

So, the displacement here is 35vdt=\int_3^5 v \, \mathrm{d}t = \cdots , where the definite integral has no constant of integration. i.e: you just integrate and then do integral(upper limit) - integral(lower limit).

The reason why this is so is, let's assume f(x)f(x) integrates to F(x)+cF(x) + c, then:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_a^b f(x) \, \mathrm{d}x = \bigg[F(x) + c\bigg]_a^b = (F(b) + c) - (F(a) +c) = F(b) - F(a) + c - c = F(b) - F(a)\end{equation*}

Reply 6
Original post by samb1234
It's always the same when you are evaluating between limits. Say you were doing [x^2 +7x +c] with bounds 4 and 0. You would get (4^2 +28 +c) -(0^2+0+c) and as you can see the c cancels


Original post by Alexion
If you've ever integrated between limits, you should be aware of this :/

If not ~ then you should find that the constant will cancel anyway.


The mark scheme didn't use limits in the integral. Im aware you don't add a constant when using limits.
Screen Shot 2016-05-22 at 13.29.58.png
Original post by ozmo19
The mark scheme didn't use limits in the integral. Im aware you don't add a constant when using limits.
Screen Shot 2016-05-22 at 13.29.58.png


yes it does. They haven't put the limits on the integral sign but what are they doing lol - they're finding the distance at the two times and subtracting them, i.e. evaluating between limits
Reply 8
Original post by ozmo19
The mark scheme didn't use limits in the integral. Im aware you don't add a constant when using limits.
Screen Shot 2016-05-22 at 13.29.58.png


The mark scheme is effectively just doing it the long way. Using limits is a very valid way to answer the question.

The way it does it effectively just breaks down the process. But if you were to do it without removing the constants, you'd get two expressions t1 + d and t2 + d so when you subtract them, they would cancel out, just like when using limits.
Reply 9
Original post by ozmo19
The mark scheme didn't use limits in the integral. Im aware you don't add a constant when using limits.
Screen Shot 2016-05-22 at 13.29.58.png


They are integrating between limits, just in a stupid and potentially confusing way.
Reply 10
Original post by Zacken
They are integrating between limits, just in a stupid and potentially confusing way.


Do i need to integrate between 3 and 4 and then between 4 and 5 then add the result since it stops when t=4 and possibly changes direction?

Thanks
Reply 11
Original post by ozmo19
Do i need to integrate between 3 and 4 and then between 4 and 5 then add the result since it stops when t=4 and possibly changes direction?

Thanks


If it changes direction, then yes.
Reply 12
Original post by Zacken
x

Why is there actually no reaction in this question????

Screen Shot 2016-05-24 at 18.06.01.png
Reply 13
Original post by ozmo19
Why is there actually no reaction in this question????

Screen Shot 2016-05-24 at 18.06.01.png


Who says there isn't one?
Reply 14
Original post by Zacken
Who says there isn't one?


I redraw a diagram and did the question with one but the mark scheme doesn't include a reaction
(edited 7 years ago)
Reply 15
Original post by ozmo19
I redraw a diagram and did the question with one but the mark scheme doesn't include a reaction


Link the markscheme?
Reply 16
Original post by Zacken
Link the markscheme?


EDIT: the ms doesn't make use of a reaction force to calculate either answers. Just to clarify, there is one and I should include one on the diagram in the exam?
Thanks
Reply 17
Original post by ozmo19
EDIT: the ms doesn't make use of a reaction force to calculate either answers. Just to clarify, there is one and I should include one on the diagram in the exam?
Thanks

MS doesnt make use of a reaction force, because you don't need it to do the question. If, in the exam, you are asked to draw out the diagram marking all forces, then include the normal reaction, yes.
Reply 18
Original post by ozmo19
EDIT: the ms doesn't make use of a reaction force to calculate either answers. Just to clarify, there is one and I should include one on the diagram in the exam?
Thanks


There is one and if asked to show all forces you wpuld need to include it. It's not required for this question because you need only resolve parallel.

Quick Reply

Latest