Help with Edexcel M1 - Exercise 3E)

Hi guys,

I've been going along with M1 recently but went on a school trip for a week and missed 5 lessons so I need to catch up but I'm struggling on 3E to grasp the main method, but up to now I've been fine (I also take physics by the way).

Here's exercise 3E for anyone who doesn't have the book:

Basically I'm stumped from the very beginning of question 1. I normally find this stuff logical but I am getting seriously confused. I've read through the examples & checked the solutions on the CD but I do not understand it. If anyone could explain me just the main ideas and a question or two I would be very grateful, thanks :biggrin:

Reply 1

generalebriety

The trick to all of these is to draw force diagrams. For example, in the first picture, there will be a force vertically downwards (weight) of magnitude 0.5g, and a force perpendicular to the plane (normal reaction). You can resolve forces - that is, provided you can find a direction in which there is no acceleration (in this case, perpendicular to the plane), you can find all the components of all the forces on the object in that direction and add them together to give you zero. A bit of simple trigonometry should tell you how to do this.

Do the examples in your book make sense now?

Reply 2

MadnessRed

There are 2 forces you need to consider. Firstly the force of gravity acting vertically.
Secondly a reaction force. Acting perpendicular to the plain.

Single alpha is the angle between the plain and the horizontal. It is also the angle between perpendicular to the plain (the reaction force) and the vertical (gravity).

By taking the 2 components of gravity, perpendicular and parralel to the plain. You can equate forces and end up with a single unresolved force parallel to the plain. (I'm making this sound a lot harder than it is :/ hopefully the diagram will explain)

Basically the reaction force counters 1 component of gravity, and the other component is responsoible for the acceleration. Then just do ma = F on that.

ma = mgsin(alpha)

Reply 3

Physics Enemy

It's all about resolving forces my friend. You want the force down the inclined plane (as in Q1), so you can then you use F = ma and get a. To get this force though, you need to resolve F = mg (that's the force due to gravity and points downwards). You use the incline angle (20 deg) to get the angle you need to resolve through.

Q1) F(Down the plane) = mgcos70

So: mgcos70 = ma
=> a = (9.81)cos70 = 3.36 ms^-2 (3 SF)

Then the Q's get harder.

Reply 4

Andy Ftw

Thanks a lot to all of you, I still struggled after reading it but I think I have the main idea now, that diagram really helped MadnessRed & all of your explanations and examples also did, thanks again guys :smile:

I'll post back if I need any more help :wink:

Reply 5

JohnDoan

Im stuck on the question number 3, i only got the slight clue of the drawing the diagram, but just cant seem to get it right, could someone please explain ? :/

Reply 6

Laura05Apple

Can anybody answer question 2? When I do R=mg4/5 I get 15 something, but the answer is 27.7N, help??

Reply 7

davros

Original post by Laura05Apple

Can anybody answer question 2? When I do R=mg4/5 I get 15 something, but the answer is 27.7N, help??

If you're resolving perpendicular to the plane you have a component of the 20N force in that direction too - i.e. 20 x (3/5) which gives you another 12 to add on :smile: