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1. 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
2. (Original post by an.p1)
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
For the last part, imagine that B has been decoupled from A so that the only force acting on it is the 200N resistive force. This is equivalent asking what happens if T = 0. Under these conditions, what will be the (negative) acceleration of B? Let's call this acceleration a2. Now, with B coupled back on to A, if the acceleration of the whole system is less than a2, the force in the coupling must be directed from A towards B (as B is now slowing down more rapidly than if it has been left on its own), i.e. the force in the coupling is compressive. You should be able to work through from this limiting value of the acceleration of the whole system to find the corresponding value of X.
3. (Original post by old_engineer)
For the last part, imagine that B has been decoupled from A so that the only force acting on it is the 200N resistive force. This is equivalent asking what happens if T = 0. Under these conditions, what will be the (negative) acceleration of B? Let's call this acceleration a2. Now, with B coupled back on to A, if the acceleration of the whole system is less than a2, the force in the coupling must be directed from A towards B (as B is now slowing down more rapidly than if it has been left on its own), i.e. the force in the coupling is compressive. You should be able to work through from this limiting value of the acceleration of the whole system to find the corresponding value of X.
can u show me the working(last part only)? that will be highly appreciated
the answer for the last part is x<60
thankyou
4. (Original post by an.p1)
can u show me the working(last part only)? that will be highly appreciated
the answer for the last part is x<60
thankyou
Can I suggest you try it yourself? First, what would the acceleration of B be if the only force acting on it was the 200N resistive force?
5. I have tried a lot.....
i guess the acceleration for b will be..
T+200=40000a
or T-200=4000a
6. X+T-360-T-200=(6000+4000)a ...( tensions cancel out)
or
X+T-360+T-200=(6000+4000)a (do not cancel out)
7. (Original post by an.p1)
I have tried a lot.....
i guess the acceleration for b will be..
T+200=40000a
or T-200=4000a
For now we’re looking at the case where T = 0
8. (Original post by old_engineer)
For now we’re looking at the case where T = 0
okay......so for b
x-200=4000a
or
x+200=4000a
9. If the ONLY force acting on B is the 200N resistive force then -200 = 4000a.

If the acceleration of the whole system is less (more negative) that the value calculated above, then there must be a non-zero T tending to retard B and the forces in the coupling are then compressive.
10. (Original post by old_engineer)
If the ONLY force acting on B is the 200N resistive force then -200 = 4000a.

If the acceleration of the whole system is less (more negative) that the value calculated above, then there must be a non-zero T tending to retard B and the forces in the coupling are then compressive.
so am i doing right?
a=-0.05

x<10000a+560 (when tensions cancel out)
x<-500+560
x<60
11. so am i doing right?
a=-0.05

x<10000a+560 (when tensions cancel out)
x<-500+560
x<60
12. (Original post by an.p1)
so am i doing right?
a=-0.05

x<10000a+560 (when tensions cancel out)
x<-500+560
x<60
Looks good to me

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