OCR M1 January 2008 Question 6iii

Hey, the question is as follows: "(iii) Calculate the magnitude of the frictional force acting on the block when the 4.9 N force acts at
an angle of 30◦ to the upward vertical, justifying your answer fully. [4]"

(weight of the block is 14.7N and the block is on a horizontal plane, mu is 1/3).

I've worked out what F would be in the horizontal (as it does in the mark scheme despite not including any acceleration the block might have) and vertical (doesn't say if it's being lifted or stationary so could have an acceleration also), it says in the mark scheme to compare the two answers and that the smallest one is correct without explanation.

Hope I've been clear enough to be understood, thanks.

Question paper: http://pmt.physicsandmathstutor.com/download/Maths/A-level/M1/Papers-OCR/January%202008%20QP%20-%20M1%20OCR.pdf
Marck scheme: http://pmt.physicsandmathstutor.com/download/Maths/A-level/M1/Papers-OCR/January%202008%20MS%20-%20M1%20OCR.pdf

That's exactly the point - we don;t know whether limiting friction has been reached or not.

So they have calculated limiting friction and then the frictional force horizontally which opposes the motion.

Since the second is smaller the block isn't moving and friction is that value.

You need to remember that there is no friction with no force then as a force is applied the friction opposes the motion until it reaches limiting value.

Ohh, so as the limiting friction is not met the block is not moving which thus states that's the friction.

I think many teachers gloss over this idea - friction <= limiting value .

I teach it using equipment so students see the forces increasng until the block starts to move ...

You've cleared the issue up well - thank you!