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M1 Exam Questions HELP!!!

Hi, need a hand with this plz if you could guide me through, cheers:

A loaded skip of total mass 2000kg is at rest on a path inclined at 21degrees to the horizontal. A force of magnitude 2500N acts upwards on the skip in a direction parallel to a line of greatest slope of the path (e.g. parallel to the slope).

i) Find the normal force exerted by the path on the skip (done this need to check answer)?

ii) Find the frictional force exerted on the skip? HELP!

iii) The force of magnitude 2500N is now removed and the skip starts to slide downwards. Given that the coefficient of friction between the skip and the path is a third (1/3), and that air resistance can be ignored, find the acceleration of the skip. Give your answer in m/s^2, correct to 2 d.p.

And this final one:

A particle P is attached to two points A and B by two light strings. P hangs in equilibrium with PA and PB making angles of 55 degrees and 25 degrees respectably with the upwards vertical, given that the tension in PA is 10N, find:

i) The tension in PB

ii) The weight of P

Cheers

Streety
streetyfatb
Hi, need a hand with this plz if you could guide me through, cheers:

A loaded skip of total mass 2000kg is at rest on a path inclined at 21degrees to the horizontal. A force of magnitude 2500N acts upwards on the skip in a direction parallel to a line of greatest slope of the path (e.g. parallel to the slope).

i) Find the normal force exerted by the path on the skip (done this need to check answer)?

R= 2000gcos21 = 18.3kN

ii) Find the frictional force exerted on the skip? HELP!

F + 2500=2000gsin21
F = 4.52kN


iii) The force of magnitude 2500N is now removed and the skip starts to slide downwards. Given that the coefficient of friction between the skip and the path is a third (1/3), and that air resistance can be ignored, find the acceleration of the skip. Give your answer in m/s^2, correct to 2 d.p.

F=ma
2000gsin21-1/3(2000gcos21) = 2000a
a=0.46ms^-2


And this final one:

A particle P is attached to two points A and B by two light strings. P hangs in equilibrium with PA and PB making angles of 55 degrees and 25 degrees respectably with the upwards vertical, given that the tension in PA is 10N, find:

i) The tension in PB

equate the horizontal components:
PAsin55 = PBsin25
10sin55=PBsin25
PB = 19.38N


ii) The weight of P

equate vertical components:
PAcos55 + PBcos25 = mg
mg=23.3N

Cheers

Streety



Hope that's all ok - I've probably made mistakes.
Reply 2
ii) Find the frictional force exerted on the skip? HELP!

F + 2500=2000gsin21
F = 4.52kN




is it definetley F+2500=2000gsin21

and why?
Reply 3
chats
ii) Find the frictional force exerted on the skip? HELP!
F + 2500=2000gsin21
F = 4.52kN
is it definetley F+2500=2000gsin21
and why?


Because the forces acting up (parallel to) the plane are equal to the forces acting down in the same direction. As it is on the verge of slipping down:
Fr + 2500 act up
2000gsin21 acts down

Hence Fr + 2500 = 2000gsin21.
Reply 4
Gaz031
Because the forces acting up (parallel to) the plane are equal to the forces acting down in the same direction. As it is on the verge of slipping down:
Fr + 2500 act up
2000gsin21 acts down

Hence Fr + 2500 = 2000gsin21.


why does friction act up? :redface:
Reply 5
Gaz031
Because the forces acting up (parallel to) the plane are equal to the forces acting down in the same direction. As it is on the verge of slipping down:
Fr + 2500 act up
2000gsin21 acts down

Hence Fr + 2500 = 2000gsin21.


Edit: I'm curious though, on the first part. If you resolve vertically you don't seem to get the same answer for the normal reaction:
Rsin69 + 2500sin21 = 2000g
R = 20KN.
I know there's a flaw in that somewhere as you do it the exact same way i always do it. What is it?
You should have done it as cos21, not sin69... if that makes a difference... Friction is always cos of the inclined angle.
Reply 6
chats
why does friction act up? :redface:

Because the skip is on the point of moving down. Friction always acts to oppose motion - so the direction of friction may change depending on the direction of motion or brink of motion of the particle.
Reply 7
RobbieC
You should have done it as cos21, not sin69... if that makes a difference... Friction is always cos of the inclined angle.


Cos21 = sin69 as cos(90-angle) = sin(angle) etc.
I sketched a dotted line across the horizontal plane where the particle is and took my angles that way.

Edit: The cos21=sin69 doesn't make a difference but friction is the thing i was missing out, it has some vertical element. o_o. I should be used to getting up early by now.
Reply 8
Gaz031
Because the skip is on the point of moving down. Friction always acts to oppose motion - so the direction of friction may change depending on the direction of motion or brink of motion of the particle.


get ya
Reply 9
chats
get ya


When are the P1/M1/S1 exams? This week or next?
Reply 10
Gaz031
When are the P1/M1/S1 exams? This week or next?


today in about 1.5 hours :biggrin:
Reply 11
chats
today in about 1.5 hours :biggrin:

Best of luck =) Anything you're unsure about that needs a little clarification? I'm sure you're going to do excellent.
I wish i could do some exams this week - to get some over with. It's going to be a pain having 7 maths exams in June and 3 in January.
Reply 12
Gaz031
Best of luck =) Anything you're unsure about that needs a little clarification? I'm sure you're going to do excellent.
I wish i could do some exams this week - to get some over with. It's going to be a pain having 7 maths exams in June and 3 in January.


thanks man, I think I know everything i need to know now.

woah 7 maths exams :eek:

good luck for january and june
Reply 13
chats
thanks man, I think I know everything i need to know now.

woah 7 maths exams :eek:

good luck for january and june


Thanks =) I'll think of the people sat doing exams while i'm there 'teaching myself' in maths 'lessons'.