# represents points as curve

• Mar 25th 2007, 05:04 AM
zak_killcity14@hotmail.co
represents points as curve
the following table gives the force F newtons which, when applied to a lifting machine overcomes a corresponding load of L newtons.

Force F newtons 25 47 64 120 149 187
Load L newtons 50 140 210 430 550 700

choose suitable scales and plot a graph of F (vertically) against L (horizontally) Draw the best straight line through the points. Determine from the graph (a) the gradient (b) the F-axis intercept (c) the equation of the graph (d) the force applied when the load is 310N (e) the load that a force of 160N will overcome (f) if the graph were to continue in the same manner, what value of force would be needed to overcome a 800N load?

this is how i have plotted the graph

http://i111.photobucket.com/albums/n...v_k/img001.jpg

from this i have worked out questions (a)(b)(d)(e)

(a)0.6896551724
(b)12.5 Newton's
(d)84N
(e)637.5N

if you think that any of these are wrong please tell me

i need help on (c) & (f)
• Mar 25th 2007, 09:53 AM
Jhevon
Quote:

Originally Posted by zak_killcity14@hotmail.co
the following table gives the force F newtons which, when applied to a lifting machine overcomes a corresponding load of L newtons.

Force F newtons 25 47 64 120 149 187
Load L newtons 50 140 210 430 550 700

choose suitable scales and plot a graph of F (vertically) against L (horizontally) Draw the best straight line through the points. Determine from the graph (a) the gradient (b) the F-axis intercept (c) the equation of the graph (d) the force applied when the load is 310N (e) the load that a force of 160N will overcome (f) if the graph were to continue in the same manner, what value of force would be needed to overcome a 800N load?

this is how i have plotted the graph

http://i111.photobucket.com/albums/n...v_k/img001.jpg

from this i have worked out questions (a)(b)(d)(e)

(a)0.6896551724
(b)12.5 Newton's
(d)84N
(e)637.5N

if you think that any of these are wrong please tell me

i need help on (c) & (f)

once you get the answer to (c), the answer to (f) will follow.

(c) asks for the equation of the line. the equation of a line is of the form:

y = mx + b
where m is the slope (or gradient) and b is the y intercept.

in this case, your y is F and your x is L, so your equation will look like

F = mL + b

leave in F and L and plug in the numbers you found for the gradient and intercept. i never really checked this, but according to you, your slope was 0.6896551724 which you found in (a) and your intercept was 12.5 which you found in (b). so the equation of your line is:

F = 0.6896551724L + 12.5 ............i don't know if you want to round off m a bit, depends on how accurate you want to be.

(f) asks what would F be if L was 800.

now that you have the formula, just plug in

when L = 800
F = 0.6896551724(800) + 12.5 = 564.2241376 N approximately

which seems a bit high to me judginf from th graph, but it's right provided you got (a) and (b) correct
• Mar 25th 2007, 09:57 AM
Jhevon
so yeah, i checked a few numbers, and i keep getting around .24 for the gradient. check your work again. i'm going to roughly check what the intercept should be

i keep getting just over 13 for the intercept, so i guess you're ok there. but double check to make sure, and see if you don't get a number above 13 rather than 12.5
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so using my approximations (which are not totally accurate becuase, like i said, i only checked the first few numbers), your answers should be close to the following:

(c) F = 0.24L + 13

(f) when L = 800

F = 0.24(800) + 13 = 205 ...........and this number makes a lot more sense to me than 500 and something--a lot more sense. just look at how your graph is sloping. at 750 you were just under 200, there's no way going just 50 units more on the L-axis you could jump more than twice that

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By the way, you should use graphing paper when drawing graphs (that you want to be accurate) by hand. chances are small that you'll get a very good answer graphing on paper like that.