Confused by Mechanics(Fmax)

I am confused by Question 8 part c-I calculated a and b correctly but part c really confuses me-now that the force of 36N has been removed then the reaction force should be 4*9.8*cos30 and the mark scheme agrees with me but the only force acting towards motion on the object is 4*9.8*sin30 and Fmax=Ru I have already correctly calculated u=0.73 so Fmax=(0.73*4*9.8*cos30)=24.7821829547N
4*9.8*sin30=19.6N so the particle shouldn't even be moving :confused:

I don't understand the mark schemes maths as well.

QP-http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202012%20-%20QP/6677_01_que_20120307.pdf
MS-http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202012%20-%20MS/6677_01_msc_20120214.pdf

(edited 10 years ago)

Reply 1

Liamnut

Original post by Dalek1099

R changes because of the 36N being removed, which means the frictional force (f=uR) changes too. I wasn't sure whether or not you'd taken that into account because I don't understand the last half of your post.

After this you need to use suvat, you know u and v and you can get a by resolving up the plane. To do this, you need to use R and to find this, you need to resolve perpendicular to the plane.

Edit: This should help: http://examsolutions.net/a-level-maths-papers/worked-solution/worked-solution.php?paper_id=56&solution=8.3

(edited 10 years ago)

Reply 2

Dalek1099

Original post by Liamnut

R changes because of the 36N being removed, which means the frictional force changes too. I wasn't sure whether or not you'd taken that into account because I don't understand the last half of your post.

I understand that the frictional force changes due to R because Fmax=uR so because u remains constant any change in R will change Fmax and have calculated the maximum friction as more than the force causing the object to move, which means the object shouldn't be moving :confused:

Reply 3

hello calum

Original post by Dalek1099

This is the question I forgot to answer in my first M1 exam! I had to retake :L I just didn't see it!
You're forgetting that the particle is already moving at 16m/s. So find the difference in the Fmax and the force acting against it. then work out the deceleration :smile:

(edited 10 years ago)

Reply 4

Felix Felicis

Original post by Dalek1099

OP, the particle continues to move up the plane rather than stopping instantaneously as the force is removed and moving downwards because, as you say, the force due to weight parallel to the plane is less than the limiting friction.

(edited 10 years ago)

Reply 5

Liamnut

Original post by Dalek1099

You have to assume the object is moving at 16ms-1 before you start the question. You're logic is kind of right, because the particle will now start to decelerate. The particle has momentum from before the 36N force was removed.

Reply 6

Dalek1099

Original post by hello calum

This is the question I forgot to answer in my first M1 exam! I had to retake :L I just didn't see it!

I have done that and have still not got the correct answer.Fmax=0.73*4*9.8*cos30=24.7821829547N and the force causing the motion is 4*9.8*sin30=19.6 Resultant F=19.6-24.7821829547=-5.1821829547=ma=4a a=-5.1821829547/4=-1.29554573868ms^-2 the markscheme says a=-11.06 so I have gone wrong again :angry:

Reply 7

Liamnut

Original post by Dalek1099

To get a,

Resolve up the plane to get the equation:

-uR-4g(sin30)=4a

Are you assuming uR and 4g(sin30) are positive? If so, this is wrong. I stated why earlier. (You have to resolve in the direction of motion and the particle will be moving

(edited 10 years ago)

Reply 8

Dalek1099

Original post by Liamnut

To get a,

Resolve up the plane to get the equation:

-uR-4g(sin30)=4a

Are you assuming uR and 4g(sin30) are possitive? If so, this is wrong. I stated why earlier.

I think I get it now am I right in thinking that the particle decelerates up the plane coming to a stop at the top of the plane because I thought the particle would start falling down the plane-if so it now makes some sense but why is friction causing motion :confused:

The friction should swap sides straight away :confused:

Reply 9

Liamnut

Original post by Dalek1099

The friction should swap sides straight away :confused:

The friction isn't causing motion, it's opposing motion, hence the friction being negative in the equation I gave you. The particle has momentum from before the 36N force was removed.

Yes, what's in bold is correct. :smile:

(edited 10 years ago)

Reply 10

Dalek1099

Original post by Liamnut

The friction isn't causing motion, hence the friction being negative. The particle has momentum from before the 36N force was removed.

Yes, this is correct :smile:

I full understand it now Thanks because the particle still continues to move up the plane but at a slower velocity then friction remains on the same side to cause this decrease in velocity. The problem was that I foolishly thought that the particle would instantly go from 16 to 0 and fall down the plane :redface: