Two forces, (4i-5j)N and (pi+pj)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), show that '2p + q + 3 = 0'
I completely don't understand what to do or where to begin, can someone show me the steps to this question please
On an unrelated matter to that question, when I find the 'deceleration' of a particle, do I put the figure as a negative number and when I use it in one of the 'S U V A T' equations do I also put it in as a negative?
x Turn on thread page Beta
need help with this M1 question, so confused watch
- Thread Starter
- 14-01-2010 21:23
- 14-01-2010 21:31
This Question is pure evil! I did it earlier today, I'll try to explain.
because it is parallel to that, but you dont know the length, we will call the length N,
(4i-5j)+(pi+qj) = Ni-2Nj
Then split to i and j, then similtaneous equation N out.
Hope that Helps
- 14-01-2010 21:44
in general the gradient is rise/run.
You know it's parallel to (i-2j) so the gradient of that is -2/1.
Equating the the forces in terms of i and j gives (4+p)i and (q-5)j.
gradient is rise/run (j/i) therfore q-5/4+p=-2 (gradient of the parallel vector)