M1 question

Two trucks A and B, of masses 6000kg and 4000kg respectively, are connected by a horizontal coupling. An engine pulls the trucks along a straight horizontal track, exerting a constant horizontal force of magnitude X newtons on truck A. The resistance to motion for truck A may be modeled by a constant horizontal force of magnitude 360N; for truck B the resistance may be modeled by a constant horiztonal force of 240N. Given that the tension in the coupling is T newtons and that the acceleration of the trucks is ams^-2, show that

T=\frac{2}{5}X

, and express a in terms of X.

Given that the trucks are slowing down, obtain an inequality satisfied by X.

The model is changed so that the resistance for truck B is modeled by a constant force of magnitude 200N. The resistance for truck A remains unchanged. For this changed model find the range of possible values of X for which the force in the coupling is compressive (i.e. the force in the coupling acting on B is directed from A to B).

How is the third part done? I did the rest. I thought it is compressive if the system is slowing down so don't we just put X less than the sum of the resistances? But this doesn't give the correct answer.

Also, on a related note, how does one handle questions with rings on wires and stuff? I can't do any of them.

Reply 1

gangsta316

Can anyone help?

Reply 2

meatball893

Horribly unclear diagram:

B---------------A---------Truck
<-240-TN <-360+TN X N->

Consider the motion of A:

X-360-T=6000a

Consider the motion of B:

T-240=4000a

Add them:

X-240-360=10,000a

X-600=10,000a

a=(X-600)/(10,000)

Sub into a previous equation...

T=4000a+240=4000(X-600)/10000 +240= (4000/10000)X -(4000*600/10000)+240=(2/5)X

I have no idea why i didn't use Latex...just silly, really...

Reply 3

meatball893

If the trucks are slowing down, then a is -ve.

\therefore \frac{X-600}{10000}<0 \implies X-600<0 \implies X<600

Much easier to read! :biggrin:

Reply 4

meatball893

ah, damn, just read the end of your post

sorry!

Reply 5

meatball893

B------------------A----------Truck
<-200N-T X-T-360->

A:

X-T-360=4000a

-200+T=6000a

All:

X-560=10000a \\ \\a=\displaystyle\frac{X-560}{10000}

T=6000a+200=\frac{6000(X-560)}{10000}+200=\frac 35 X -136

But T is -ve, as it is compressive...

\frac 35 X -136<0 \implies \frac 35 X<136 \implies X<\frac{680}3

I have no idea if I've done this right, so please don't put too much faith in it (unless it matches the answer you've been given). I hope it's right, or I haven't been much help...

Reply 6

gangsta316

I got X < 600 at first but I think the book said X < 60...

Reply 7

agbotse.joana

You made some mistakes with your calculations:
B: T = 6000a + 200 = (6000(X - 560))/10000 + 200 = (3/5)X - 336
3/5X - 336 < 0
3/5X < 336
Hence X < 560 not 680/3.

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