There was this question on my M1 exam today which I remembered perfectly and was wondering if somebody could please check it.
The question went along the lines of two barges on a slope inclined at 16.6 degrees. They are attached by a light coupling and barge A has weight 1.5 Kg and B has weight 2 Kg with resistance on one barge being 0.7 N and on the other 2.3N and there is a 10 N force up the slope acting on the front barge and I have to work out the acceleration.
Would it be : 10-0.7-2.3 -1.5gsin16.6 -2gsin16.6= 3.5a
And after a bit of rearranging I would get the answer. However I remember getting a negative accelaration and if thats true and the barges are moving down the slope then wouldn't the frictional resistances be acting in the opposite direction and at which point I get confused. Can someone just confirm my method please.
M1 Question Watch
- Thread Starter
- 27-01-2010 18:27
- 27-01-2010 23:41
Hey Unknown, I do have something to say to help you, but when exactly was this exam, as you should be aware exam discussion is prohibited untill a certain time after the exam has finished
- 28-01-2010 03:05
What exam board was it? Certain exam boards can be discussed straight after the exam. Were the barges initially at rest, or didn't it specify?
- 28-01-2010 03:13
I just checked online - the only M1 exam on the 27th was the OCR MEI paper, so that's ok for discussion (I believe it's only Edexcel that's barred from discussion).
I'm rubbish at Mechanics so can't help you... Sorry
- 28-01-2010 03:39
Thanks lareneg for the check ;D.
SO to Unknown? Well yes from the info that you have given, yes it would seem that the barges would be sliping down the slope. That obviously then doesnt make and awful lot of sense as you said the frictional resistance would be acting in the other direction.
That said however, you initialy said just resistance, not 'frictional' resistance, so you could say then, barges to me implies river barges, and so resistance could be say the water current which would always act in the same direction regardless of direction of motion, all resistance doesnt have to be frictional ;D.
Other than that I can only say pehaps you didnt remember it as perfectly as you thought. As musicbaserich pehaps they wernt initaly at rest, although that wouldnt have affected the acceleration on the bodies it would mean that a 'Frictional Resistance' would be acting down the incline. Or simply coefficients changed here and there and the problem would resolve a possitive acceleration up the slope
- 29-01-2010 04:29
I also sat this M1 paper and have argued with some other people as to the answer to this question, and many people have many answers.
Im fairly sure that the question did mention friction and not just a resistance.
There is the 10N pulling both the barges up the slope.
The mass of the two barges creates a force of 9.8N down the slope
( (1.5+2)x9.8x(2/7) )
The value of 2/7 is what was provided, Sin(alpha)=2/7
10N - 9.8N leaves a value of 0.2N resultant force to pull the barges up the slope, however the frictional forces then prevent the barges for moving. A lot of people I talked to simple minused the frictional forces, forgetting that that is what they are and do not cause motion, their answer then showed the barges traveling down the slope.
Unfortunatly out of the people I have spoken to, I have not come across a person with the same answer, nor have I come across a person who has managed to justify their answer.
I hope that helped.