M1 May 2008 (examiner or me?)
Okay so I was working through this paper and was doing fine until I marked my answer in question 5b, apparently the examiners think its meant to be
( X - 15cos30 = Rcos50) instead of (Rcos50 + X - 15cos30 = 0).
Can any kind soul please please tell me that I resolved the forces correctly, I am on so much pressure cracking my brain here to figure out why they think its the first one.
thankyou
( X - 15cos30 = Rcos50) instead of (Rcos50 + X - 15cos30 = 0).
Can any kind soul please please tell me that I resolved the forces correctly, I am on so much pressure cracking my brain here to figure out why they think its the first one.
thankyou
Rcos50 is the resultant force of P and Q. So if you take Q going to the right and minus P going to the left you get the resultant force.
Don't think of R as another force going to the right that is acting with the other forces going to the right that must be balanced out with the left: it is the resultant force.
In class we did this a different way though:
You can form a triangle OQR and use sine rule to find both R and X.
Don't think of R as another force going to the right that is acting with the other forces going to the right that must be balanced out with the left: it is the resultant force.
In class we did this a different way though:
You can form a triangle OQR and use sine rule to find both R and X.
(edited 13 years ago)
Original post by Thrug
Rcos50 is the resultant force of P and Q. So if you take Q going to the right and minus P going to the left you get the resultant force.
Don't think of R as another force going to the right that is acting with the other forces going to the right that must be balanced out with the left: it is the resultant force.
In class we did this a different way though:
You can form a triangle OQR and use sine rule to find both R and X.
Don't think of R as another force going to the right that is acting with the other forces going to the right that must be balanced out with the left: it is the resultant force.
In class we did this a different way though:
You can form a triangle OQR and use sine rule to find both R and X.
Dang!, should've have stayed on this. Sorry for late reply, I kinda gave hope anyone was going to bother just getting out of bed to resolve mechanics.
So in essence, Q - P = R then. Makes perfect sense. I thought of the triangular method but seems to me I prefer this one.
Original post by mauvetard
Dang!, should've have stayed on this. Sorry for late reply, I kinda gave hope anyone was going to bother just getting out of bed to resolve mechanics.
So in essence, Q - P = R then. Makes perfect sense. I thought of the triangular method but seems to me I prefer this one.
So in essence, Q - P = R then. Makes perfect sense. I thought of the triangular method but seems to me I prefer this one.
Yea that's the first time I've seen the forces resolved into a triangle so I'd just stick to resolving how I normally do it.
And yes, as always R is the combination of the other forces:
R = Q + P. But as Q and P are in opposite directions, it's kinda like Q + -P = R.
Original post by Thrug
Yea that's the first time I've seen the forces resolved into a triangle so I'd just stick to resolving how I normally do it.
And yes, as always R is the combination of the other forces:
R = Q + P. But as Q and P are in opposite directions, it's kinda like Q + -P = R.
And yes, as always R is the combination of the other forces:
R = Q + P. But as Q and P are in opposite directions, it's kinda like Q + -P = R.
Agreed!. Now Im sure as to not make a stupid mistake in the exam, wouldn't want to meddle with too many methods
.Thanks for clarification!.Quick Reply
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