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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

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

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.

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?

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.

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?

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?

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.

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.

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

good luck for january and june

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