M1 Edexcel - "Dynamics of a particle on an inclined plane" question

The 20N force is not parallel to the plane, so there's obviously going to be a component included. Just resolve perpendicular to the plane, including the weight, 20N force and R (reaction force), and the sum of the three forces must be zero due to there being no acceleration in that plane.

R + mg + 20N = 0
R + 19.6 + 20 = 0
R = -39.6?

Erm, not quite. The key word was resolve.

R is the only force that is acting perpendicular to the plane, both the weight and the horizontal forces are not.

how do i resolve with that 20N force?

do i make a seperate triangle with mg going straight down and 20N going to the right? because the hypoteneuse created is the opposite direction of the perpendicular force

The 20N force is horizontal, you know the angle between the horizontal force and the slope, it's the same as the angle the slope is inclined to. You want to resolve it so that it's perpendicular, so it would be 20sin(x) (where x is the angle the slope is inclined to).

The weight is directly down, so creates a triangle in itself. The weight must be (90-x) degrees from the slope. So when resolving perpendicular to the slope you get sin(90-x) which is equivalent to cos(x)

R - 20sinx - 2gcosx = 0

wtf.. the triangle is a right angle, why does mg make a (90-x) degree angle on the slope if 20N is horizontal?

Thanks for the PM.

The calculation you done is absolutely correct as far as I can tell in the situation where the 20N force doesn't act on the box. However lets now take in to account the 20N force in our calculations in order to find the normal reaction, R. In order to do this we resolve the forces perpendicular and parallel to the plane;

f=ma, on the 2kg box, taking down the slope as the positive direction.

-20cos(arctan(3/4)) + 2gsin(arctan(3/4)) + Rsin(arctan(3/4)) = 0 as no acceleration so ma=0 (right hand side). You know everything in here apart from R, so rearrange it and find R.

The weight has absolutely nothing to do with the 20N horizontal force. No offence, but the questions you're asking are the fundamental basics of M1. It'll be a lot easier if you go back over the book and make sure you fully understand it, as opposed to me trying to explain what the book has already said.

Thanks for the PM.

The calculation you done is absolutely correct as far as I can tell in the situation where the 20N force doesn't act on the box. However lets now take in to account the 20N force in our calculations in order to find the normal reaction, R. In order to do this we resolve the forces perpendicular and parallel to the plane;

f=ma, on the 2kg box, taking down the slope as the positive direction.

-20cos(arctan(3/4)) + 2gsin(arctan(3/4)) + Rsin(arctan(3/4)) = 0 as no acceleration so ma=0 (right hand side). You know everything in here apart from R, so rearrange it and find R.

i have the book in front of me and it doesn't explain sh!t. i would have finished M1 months ago if this part of the chapter was explained properly.

Which book?

hi, thanks for the help.

what i still don't understand is what triangle u make to obtain the values like -20cos(arctan(3/4)) and 2gsin(arctan(3/4)) etc.

can u tell me what the sides of the triangle are, i'm sure i'll understand everything from there. thanks

You can't resolve R to the line of slope because the angle it always makes with the slope is 90 degrees and cos90 = 0.

OK the triangles are made as follows...

-20cos(arctan(3/4) : The minus is there because we said that down the slope is the positive direction and the component of the 20N force acting parallel to the plane is up the slope (pushing the box up the slope). Now, the 20N force acts horizontally, make the 20N vector (or 'line') the hypotheneuse (yeah spelling I know!) and make a right angles triangle using 2 other sides linked to it.

why is 20N the hypotenuse if it acts horizontally?