m1 june 2002 paper

im stuck on m1 june 2002 EDEXCEL question 4
to do with a horizontal force acting on P

how do i find the reaction force?
i got the mark scheme but it makes no sense!!
can sum1 if possible even try and include a diagram

id be really grateful!!

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Reply 1

phys981

Can you give me the question? I assume you mean edexcel because AQA m1 June 2003 has nothing to do with this....

Reply 2

ash213

its too long to write
sorry , dont mean to be fussy!

Reply 3

insparato

A box of mass 6 kg lies on a rough plane inclined at an angle of 30 to the horizontal. The box is held in equilibrium by means of a horizontal force of magnitude P newtons, as shown in Fig. 2. The line of action of the force is in the same vertical plane as a line of greatest slope of the plane. The coefficient of friction between the box and the plane is 0.4. The box is modelled as a particle.

Given that the box is in limiting equilibrium and on the point of moving up the plane,
find,

(a) the normal reaction exerted on the box by the plane,

(b) the value of P.
The horizontal force is removed.

(c) Show that the box will now start to move down the plane.

Christ sake.

I had trouble with this, the mark scheme just doesnt make sense.

Reply 4

uk6458

rsk, i found their question, i have attached it....... :smile:

damm only just realised you typed it up insparato :redface:

question.JPG46.2KB

Reply 5

phys981

Well, I have something, but they ask you to find R first then P, and I can only see how to find P first, then R; so must be doing something wrong.

Reply 6

phys981

Sorry, brain is fuddled. Will have another look in the morning. Meantime, if you post the answers the markscheme gives that might help.

does any1 kno???

mmm I haven't done M1 in a while.. but maybe they just want the normal in terms of P

So you say:

a) R = mg cos30 + P sin30

b) Resolving parallel to the plain:

mg sin30 + Fr = P cos30
mg sin30 + "mew" R = P cos30
mg sin30 + 0.4(mg cos30 + P sin30) = P cos30
mg sin30 + 0.4 mg cos30 = P cos30 - 0.4 P sin30
mg sin30 + 0.4 mg cos30 = P(cos30 - 0.4 sin30)
P = (mg sin30 + 0.4 mg cos30)/(cos30 - 0.4 sin30)

c) P removed, so:

R = mg cos30
Fr = 0.4 x mg cos30
But mg sin30 > 0.4 mg cos30
So it will move down the plane..

I dunno if this is all right though... what does the mark scheme say?

Reply 9

phys981

The only way I can see to find R without knowing P is to resolve horizontally and vertically (something I thought you were told never to do!) instead of perp and parallel.

Doing this gives me 55.2 for R and 46.7 for P. Thoughts, anyone?

However, since the original poster couldn´t even be @rsed to post the Q

its too long to write
sorry , dont mean to be fussy!

I don´t really know why I´m bothering.....

Reply 10

insparato

The mark schemes i have just dont make sense for some of the papers. Its not really doing good for my confidence anyways here it is.

Its a funny filename but its the mark scheme. I think R = 88.2N if i recall.

Reply 11

phys981

88.2 sounds like what I got last night doing it by perp and parallel - but hat to get P first as i recall - will have a look at post back

Reply 12

Kathrin

did it - fairly easy!

for a)

6g = R cos30° - Fr cos60° (1)

we know Fr = mu x R --> Fr = 0.4 x R (2)

(2) in (1)

6g = R cos30° - 0.4 x R cos60°
6g = R x (cos30° - 0.4 x cos60°)
R = 88.28N

for b)

Fr = P cos30° - 6g cos60°
mu x R = P cos30° - 6g cos60°
0.4 x 88.28 = P cos30° - 6g cos60°
35.31 = P cos30° - 6g cos60°
P cos30° = 64.71
P = 74.73

diagram2.GIF2.9KB

Reply 13

np_desi

ash213

lol i remember doing this yday, and yeah i got most of it...

anyway, wat they did is that they give u mu [funny U]

u know F = mu R

so F = 2/5 R

then jus resolve as u would, making P the subject of both perp and parallel to the slope. Then u equate both of the equations since they are both P =...

finally u know F = 2/5 R, so u replace F with 2/5 R, and i think that is it... u jus do the maths and get R :cool:

Reply 14

np_desi

Kathrin

did it - fairly easy!

for a)

6g = R cos30° - Fr cos60° (1)

we know Fr = mu x R --> Fr = 0.4 x R (2)

(2) in (1)

6g = R cos30° - 0.4 x R cos60°
6g = R x (cos30° - 0.4 x cos60°)

R = 88.28N

bah u beat me to it by just a few secs :biggrin:

anyway doing s2 and m1 at the same time is not easy :biggrin:

Reply 15

silent ninja

All you need to do is resolve vertically, perpendicular to P i.e use P as your x-axis. This elminates P's vertical compononents and you only have R. You are simply changing the axis for resolving.

so vertically, Rcos30 - 6g - 0.4Rsin30= 0
R(cos30 - 0.4sin30)= 6g => R= 88.3N

Reply 16

Kathrin

and for c)

force of weight
F = 6g cos60°
F = 29.4N

R = cos30°
R = 50.92N

Fr = mu x R
Fr = 0.4 x 50.92
Fr = 20.37N

so we know 29.4N > 20.37N
due, parcel moves down the slope

diagram3.GIF2.5KB

Reply 17

phys981

ok, yes, well at least I got 88.1 last night. had to rescue paper from the bin but here we go....

first I resolved parallel to plane

Downwards we have the friction and the component of the weight. Friction is 0.4R, weight is mgsin30
Upwards we have the component of P, P cos30

So

0.4R + mg sin30 = P cos30 ............................................ (1)

The resolve perpendicular

Upwards we have R , downwards we have the weight component (mgcos30) and the component of P (Psin30)

So R = mg cos30 + P sin30 .................................................(2)

Now subs this expression for R into (1)

0.4 (mg cos30 + P sin30) + mg sin30 = P cos30

0.4 mg cos30 + 0.4 P sin30 + mg sin30 = P cos30

20.4 + 0.2P + 29.4 = P cos 30

49.8 = P (cos30 - 0.2)

49.8 = P (0.866 - 0.2)

P = 49.8 / 0.666 = 74.7

Now all we have to do is subs this back in (2)

R = mg cos30 + P sin30

R = 50.9 + 37.35 = 88.25

Hurrah!

Reply 18

phys981

Amazing how many people can post in the time it takes to type something like that....