I got a math assingment to do over xmas and im stuck on the last 2 questions So i dunno whether anyone on here can help me and explain what i need to do to work them out

Heres the first question:

the law connecting friction F and load L is of the form

F = aL + b

where a and b are constants
when F = 5.6, L = 8
and
when F = 4.4, L = 2
find values for a, and b, and the value of F when L = 6.5

and the second question is:

The resistance R ohms of a length of wire at
T oC (degress C) is given by R = Ro (1 + at)
Where Ro is the resistance at 0 oC and a is the temperature coefficent of resistance in / oC

Find the values of a and Ro if R = 30 (omega symbol i think) at 50 oC and R = 35 (omega) at 100 oC

If anyway can help id be very thankful
Cheers Bob

Are you american?

I'll give you a hand.

F=aL+b

when F = 5.6, L = 8

5.6=8a+b
5.6-b=8a (i)

also when F = 4.4, L = 2

4.4=2a+b
4.4-b=2a (ii)

Equalise the values of a. Multiply equation ii by 4

5.6-b = 4(4.4-b)
5.6-b = 17.6-4b
5.6 = 17.6-4b+b
5.6 = 17.6-3b
3b = 17.6-5.6
3b = 12
b = 4

See how you get on with that help. Should be easy to find the value of a

H2DaE said:

Are you american?

Click to expand...

Damn it Hellcat you beat me to it [:yeahthat:

I was referring to "math", you missed the 's' off the end

H2DaE said:

I was referring to "math", you missed the 's' off the end

Click to expand...

thats cause i didnt know whether to put mathes or maths lol so i left it like that

Cheers Hellcat i got it now

just question 2 now

Tried the 2nd one but my division is letting me down... Something to do with never learning how to properly at primary/secondary school and bluffing my way through A level maths!

As far as I have got is

30 - 1 = 50a (i)
Ro

35 - 1 = 100a (ii)
Ro

Multiply eq. i by 2.. Now I'm stuck

Bob in disguise said:

thats cause i didnt know whether to put mathes or maths lol so i left it like that

Click to expand...

So do you need help with spelling (or should that be speling ) too?
H

hmallett said:

So do you need help with spelling (or should that be speling ) too?
H

Click to expand...

Hence me learning a trade that doesnt need spelling

Can't say fairer than that!
H

me doing engineering at uni next year.... please say i dont have to do that malrkee!! (sp?) i hate math, am crap at it

although can do it with hellcats answers

charlie

Hellcat said:

Tried the 2nd one but my division is letting me down... Something to do with never learning how to properly at primary/secondary school and bluffing my way through A level maths!

As far as I have got is

30 - 1 = 50a (i)
Ro

35 - 1 = 100a (ii)
Ro

Multiply eq. i by 2.. Now I'm stuck

Click to expand...

Put the left half of both equations in brackets. It might help.

Nope that Ro on the bottom is still doing my head in.


edit: Is the answer 25? Sounds about right.

25 ohm @ 0 degrees
30 ohm @ 50
35 ohm @ 100

Hellcat said:

Nope that Ro on the bottom is still doing my head in.


edit: Is the answer 25? Sounds about right.

25 ohm @ 0 degrees
30 ohm @ 50
35 ohm @ 100

Click to expand...

I got 25 too

Rearrange both equations to make it equal to a. Multiply eqn. i by 2. Eliminate the a terms, and subtract. You should have

2(30/Ro) - (35/Ro) = 1

Multiply through by Ro:

(2(30/Ro)*Ro) - ((35/Ro)*Ro) = (1*Ro)

The Ro terms on the left hand side cancel out giving 60-35=Ro=25

I go away for a few days and miss a bit of maths, grr

Hellcat said:

I'll give you a hand.

F=aL+b

when F = 5.6, L = 8

5.6=8a+b
5.6-b=8a (i)

also when F = 4.4, L = 2

4.4=2a+b
4.4-b=2a (ii)

Equalise the values of a. Multiply equation ii by 4

5.6-b = 4(4.4-b)
5.6-b = 17.6-4b
5.6 = 17.6-4b+b
5.6 = 17.6-3b
3b = 17.6-5.6
3b = 12
b = 4

See how you get on with that help. Should be easy to find the value of a

Click to expand...

Can you just explain why you multiply the ii equation by 4?

So that equation i and ii are equal so that they can b can be removed leaving only a to be worked out.

you've kinda lost me, why 4?

Edit: i just noticed why!

Just the second question im stuck on now