Mechanics question on magnitudes of vectors Watch

laurawoods
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But then on this question from January 2009, for the last part, where the examiners have given the acceleration as a magnitude rather than the vector, we would be forced to work in magnitudes. Isn' t it so??????????????????/


6. Two forces, (4i – 5j) N and (pi + qj) N, act on a particle P of mass m kg. The resultant of the two forces is R. Given that R acts in a direction which is parallel to the vector (i – 2j),
(a) find the angle between R and the vector j,
(b) show that 2p + q + 3 = 0.
Given also that q = 1 and that P moves with an acceleration of magnitude 8√5 m s –2
(c) find the value of m.


My question then is: Why does it not lead to the wrong answer in this particular example, when we are using magnitudes whereas it would have lead to a wrong answer in this question :


A particle P moves with constant acceleration (2i – 5j) m s –2 . At time t = 0, P has speed u m s –1 . At time t = 3 s, P has velocity (–6i + j) m s –1.
Find the value of u.
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Mr M
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I think you making things more difficult than they need to be Laura. Just always use vectors and, if you are asked to find the magnitude of a vector at some point, use Pythagoras' Theorem.

The direction of acceleration will be in the same direction as the resultant force if that is the thing confusing you because of F = ma (m is a scalar).

If you want to discuss these questions further, show your working.
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laurawoods
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(Original post by Mr M)
I think you making things more difficult than they need to be Laura. Just always use vectors and, if you are asked to find the magnitude of a vector at some point, use Pythagoras' Theorem.

The direction of acceleration will be in the same direction as the resultant force if that is the thing confusing you because of F = ma (m is a scalar).

If you want to discuss these questions further, show your working.
Okthanks ... i think i understand things now...and also a separate question to ask you....what is the sort of hardest vector question that they could ask us? Because I thought that the one with the resultant force was pretty hard and there wasn't such one example in the textbook. And also just another one question... Do we have to learn to used vector triangles?
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laurawoods
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(Original post by Mr M)
I think you making things more difficult than they need to be Laura. Just always use vectors and, if you are asked to find the magnitude of a vector at some point, use Pythagoras' Theorem.

The direction of acceleration will be in the same direction as the resultant force if that is the thing confusing you because of F = ma (m is a scalar).

If you want to discuss these questions further, show your working.

I am pretty stuck on this question :


Q4) FROM JAN 06

Two Forces P and Q act on a particle. The force P has magnitude 7N and acts due north. The resultant of P and Q is a force magnitude 10N acting in direction with beaing 120 cegrees. Find:

i) The magnitude of Q
ii) The direction of Q giving your answer as a bearing.


I looked at the mark scheme and was confused with this bit :
Well , they drew a vector triangle and then said
Q^2=10^2+7^2+2*10*7*cos60


Where did they get cos 60 from becuase according to the cosine rule, i actually thought that it was going to be cos120?
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Mr M
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(Original post by laurawoods)
I am pretty stuck on this question :


Q4) FROM JAN 06

Two Forces P and Q act on a particle. The force P has magnitude 7N and acts due north. The resultant of P and Q is a force magnitude 10N acting in direction with beaing 120 cegrees. Find:

i) The magnitude of Q
ii) The direction of Q giving your answer as a bearing.


I looked at the mark scheme and was confused with this bit :
Well , they drew a vector triangle and then said
Q^2=10^2+7^2+2*10*7*cos60


Where did they get cos 60 from becuase according to the cosine rule, i actually thought that it was going to be cos120?
If you don't like vector triangles you don't have to use them so long as you have another method you are secure with. They are likely to be the most efficient method in many cases though.

Drawing a diagram and making sure you mark arrows to indicate the directions of your vectors should explain your angle problem. Remember when vectors are arranged in a vector triangle they are placed nose to tail.

If you don't understand what I am talking about you need to see your own teacher. I would advise you to do that anyway.
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laurawoods
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(Original post by Mr M)
If you don't like vector triangles you don't have to use them so long as you have another method you are secure with. They are likely to be the most efficient method in many cases though.

Drawing a diagram and making sure you mark arrows to indicate the directions of your vectors should explain your angle problem. Remember when vectors are arranged in a vector triangle they are placed nose to tail.

If you don't understand what I am talking about you need to see your own teacher. I would advise you to do that anyway.
Hello Mr M , thank you for the help offered...

Just one more last question and I promise this is actually last....you know when there is a thrust question, e.g . with there being a thrust in the tow bar, at that time , do we put the deceleration arrow to be going backwards?
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Mr M
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(Original post by laurawoods)
Hello Mr M , thank you for the help offered...

Just one more last question and I promise this is actually last....you know when there is a thrust question, e.g . with there being a thrust in the tow bar, at that time , do we put the deceleration arrow to be going backwards?
That's a double negative. The acceleration arrow is probably going backwards but it depends on the physical situation at the time.
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laurawoods
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That's a double negative. The acceleration arrow is probably going backwards but it depends on the physical situation at the time.
ok thank you ...
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laurawoods
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(Original post by laurawoods)
ok thank you ...
so is the braking force going backwards?
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