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Author Topic: Dynamics  (Read 14559 times)  Share 

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d0minicz

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Dynamics
« on: August 10, 2009, 09:17:28 pm »
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Hey yo

Find the magnitude and direction with respect to of the resultant of the following forces attached:


thanks
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Damo17

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Re: Dynamics
« Reply #1 on: August 10, 2009, 09:36:54 pm »
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Quote from: d0minicz on August 10, 2009, 09:17:28 pm
Hey yo

Find the magnitude and direction with respect to of the resultant of the following forces attached:


thanks


move the 2N force down so the tip is touching O. The draw a line from tail of 2N force to the tip of 3N force, which we will can side b. Using cosine rule:




to find angle:
C is angle made by side b and 3N force.



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d0minicz

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Re: Dynamics
« Reply #2 on: August 10, 2009, 09:41:04 pm »
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Hey thanks alot !
Could you also use a graph to do this?
Would the position of X have any effect on that method?
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TrueTears

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Re: Dynamics
« Reply #3 on: August 10, 2009, 09:45:34 pm »
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Quote from: d0minicz on August 10, 2009, 09:41:04 pm
Hey thanks alot !
Could you also use a graph to do this?
Would the position of X have any effect on that method?
No it wouldn't.
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d0minicz

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Re: Dynamics
« Reply #4 on: August 10, 2009, 09:55:27 pm »
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alright cheers

just another confirmation

Three forces 5N , 7N and PN act on a particle at O. Find the value of P that will produce a resultant force along if the line of action of P N force is perpendicular to OX.
Confused about the direction of P
Would the P vector be -Pj ?
attached,
thanks again
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TrueTears

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Re: Dynamics
« Reply #5 on: August 10, 2009, 10:01:44 pm »
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No vector P is just Pj
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kamil9876

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Re: Dynamics
« Reply #6 on: August 10, 2009, 10:04:58 pm »
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it could be -Pj or Pj but you need this to be consistent with the other vectors. To illustrate what I mean, if you draw your y axis down. Then the force of that vector is -Pj. But then the j component of the vector point up would also be negative, and the j component of the vector pointing down would be positive. If the y axis was up, the vector in question woudl be Pj but the j component of the vector going up would be positive and the j component of the vector going down would be negative. So it all depends on how you build your axis - which is a useful method in solving complicated problems (choosing a convenient axis), it's all arbitrary as long as you have things consistent ie: have the things that go in the same direction the same sign, otherwise it's wrong.

But in this problem it doesn't matter, eitehr way works though i advise you to use y axis upwards since it's more conventional and what your used to, plus it also minimizes the number of negative numbers you are working with (1 rather than 2)
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TrueTears

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Re: Dynamics
« Reply #7 on: August 10, 2009, 10:05:37 pm »
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Quote from: kamil9876 on August 10, 2009, 10:04:58 pm
it could be -Pj or Pj but you need this to be consistent with the other vectors. To illustrate what I mean, if you draw your y axis down. Then the force of that vector is -Pj. But then the j component of the vector point up would also be negative, and the j component of the vector pointing down would be positive. If the y axis was up, the vector in question woudl be Pj but the j component of the vector going up would be positive and the j component of the vector going down would be negative. So it all depends on how you build your axis - which is a useful method in solving complicated problems (choosing a convenient axis), it's all arbitrary as long as you have things consistent ie: have the things that go in the same direction the same sign, otherwise it's wrong.
Yes, but why would you bother having the negative y axis as positive? Waste of time for this Q.
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kamil9876

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Re: Dynamics
« Reply #8 on: August 10, 2009, 10:08:09 pm »
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yeah i know just saying in general so that the OP has a feel for axis and can use them efficiently in more complicated scenarios by understanding how arbitrary assigning direction is. Didn't see your post so I was in no way attacking it or anything. Plus I agree with your convention of Pj rather than -Pj.
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kamil9876

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Re: Dynamics
« Reply #9 on: August 10, 2009, 10:30:32 pm »
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Sorry for Hijacking:

Damo's solution gave me an interesting idea:


The other way to solve that problem would be to make OX the direction of the positive x axis. Let's generalise by calling the 2N vector and the other one bi. Now essentially we want to find:



We can do this as Damo has done, to add the two vectors. The resulting vector sum would be:





However the angle between the two vectors is 180-50=130 so in order to get our expression in terms of that we just simply use and sub it into the expression to get:



but and so just sub in to get the more recognizable form of:



Which proves the cosine rule using vectors.


Moral of the story is that if you find two different ways of solving the same problem, you can equate them to find interesting proofs/relationships :). So always be on the lookout (except possibly exam time)
« Last Edit: August 10, 2009, 10:32:34 pm by kamil9876 »
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d0minicz

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Re: Dynamics
« Reply #10 on: August 11, 2009, 08:07:40 pm »
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lol nice


Find the resultant force of this using triangles of forces:

need to see the working and how to rearrange it to make a triangle

thanks
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kamil9876

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Re: Dynamics
« Reply #11 on: August 11, 2009, 08:25:19 pm »
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Of course, you can always break up into components and add the forces like vectors.

But as you ask: "slide" the other vector so that it meets the head of the other as shown below. The angle BAO is 20+30 since <CAO=20 (alternate angle) and <CAB=30(because it's just been translated, the angle it makes parralel to x axis is still 30).
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d0minicz

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Re: Dynamics
« Reply #12 on: August 12, 2009, 09:52:46 pm »
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Q. In a lift that is accelerating upwards at a spring balance shows the apparent weight of a body to be 2.5 kg wt. What would be the reading if the lift were at rest?

thanks
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Re: Dynamics
« Reply #13 on: August 12, 2009, 10:07:54 pm »
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  (resultant force must be up, so for convenience we take up to be positive)







When the lift is at rest the lift will read

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d0minicz

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Re: Dynamics
« Reply #14 on: August 12, 2009, 10:10:17 pm »
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ah thanks man
So is kg wt the same as Newtons ?
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