# Oxford PAT 2015, Question 15

## 6 thoughts on “Oxford PAT 2015, Question 15”

1. Sam says:

As the first person said, I left cos(45) as root2/2. Do you think they would negate marks for this as I got c=2a/root2 using this:/

1. You’ve still got the right answer. But it’s always a good idea to simplify your answer as much as possible, so if you end up with 2/(√2) you should normally rewrite it to √2.

2. mackliyn says:

cos 45 is root2/2 so how is the answer aroot2

1. Yes, cos 45 equals (root 2)/2 which equals 1/root 2. By resolving parallel to A, we see that A = C cos 45, so A = C / root 2, so C = A root 2.

1. rickardo says:

the first line, why is it not Ccos25-a=0 thats how forces are resolved in my m1 book on a very similar question?
because the object isnt moving

1. You mean C cos 45 – A = 0, right? (you wrote cos 25). That is basically the same as what I have put, when I have written C cos 45 + A = 0, the difference being in the sign of A. Both approaches are correct, but I think there are good reasons for doing it the way I have done it.

First, let’s just recap the bit that is the same for both solutions: because the object does not move, we know that the resultant force on it is zero; so whatever direction you resolve in, the sum of the forces on the mass must be zero; so if we add up all the forces the result must equal zero.

The bit where we differ is, whether you treat A as just a magnitude, or as a magnitude and sign. If you treat A as just a magnitude (the approach used in your book) you get C cos 45 – A = 0; I prefer to treat A as a magnitude and sign, in which case to get the resultant force you just add up all the different forces (resolved along the line), and you get C cos 45 + A = 0.

If you can’t see why this matters, don’t worry. Both approaches are correct. In a case like this it really doesn’t matter because it is so simple, but I prefer my approach for more complicated problems because I think it’s easier to control when things get more complex.