forces...

OP

how do we find what m is? :s-smilie:

Reply 3

nota bene

15

we don't need to find m, that is not part of the question, as the mass will be constant.

and I can't spot any mistake in chewwy's working right now, so don't know why it doesn't give the answer the book has...

Reply 4

nota bene

15

chewwy

weight is mg
frictional force = (mu)R

resultant force parallel to slope = F = ma = mgsina - (1/4)mgcosa =
(3/5)mg - (1/4)(4/5)mg = 2/5mg

wtf.

Is that right? or should it be 2/5?

Reply 5

OP

im guessing to get the 8 marks there needs to be a looong answer, maybe substituting or something, i just haven't figured it out yet.. :s-smilie:

Reply 6

OP

im quite confused, as the ball-bearing won't be in equilibrium, how do we find the resultant force, just from friction?

Reply 7

DFranklin

18

chewwy

weight is mg
frictional force = (mu)R

resultant force parallel to slope = F = ma = mgsina - (1/4)mgcosa =
(3/5)mg - (1/4)(4/5)mg = 2/5mg

wtf.

The ball isn't starting from rest, it's already moving. So both gravity and friction are working to slow it down.

So you have (3/5)mg+(1/4)(4/5)mg = 4/5mg.

Reply 8

datr

12

chewwy

weight is mg
frictional force = (mu)R

resultant force parallel to slope = F = ma = mgsina - (1/4)mgcosa =
(3/5)mg - (1/4)(4/5)mg = 2/5mg

wtf.

If the machine is firing it up the slope wouldn't friction and weight be working against the bb? I.e.

resultant force parallel to slope = (3/5)mg + (1/4)(4/5)mg = (4/5)mg

F = ma = (4/5)mg => a = (4/5)g

Reply 9

OP

ah, that works now!! Thank you all soo much! :smile:

but I still don't understand why there is 8 marks for it.. :s-smilie:

Reply 10

OP

what does the question mean by state its direction, isn't it just up the plane anyway, as the machines fires them that way?

Reply 11

Chewwy

17

no. weight and friction are pulling it down the plane.

Reply 12

No, the acceleration goes down the plane. Otherwise it'd speed up going up the plane, but it slows down. :wink:

Gah, I read the first few posts and tried it myself and got 2/5g too. :confused:

My mechanics has gone downhill since starting M4, if you'll excuse the pun.

Reply 13

OP

lol @ the pun.. :biggrin:

after reading the posts i was quite confused about how to get the force when the bb wasn't in equilibrium, but i guess it triggered something because i got it the next time i tried.. :smile:

thanks to all once again.. :smile:

generalebriety, how can you tell that its retardation, is it because acceleration is 4/5g, so you have to multiply it with -9.8, which makes it negative? :s-smilie:

Reply 14

OP

ah, the question gets annoying as I move on..

b) Given that the machine is placed at a point A, 30m from the top edge of the plane, and the ball-bearings are projected with an initial speed of 20ms-1,
find, giving your answer to the nearest cm, how close the bb get to the top edge of the plane?

do we use equations of motion, u=20, a= 7.84, s=?, t=?, v=?

great, i only know acceleration and initial speed, how do i get the max distance the bb travels? :s-smilie:

Thanks in advance.. :smile:

Reply 15

n0b0dy

lol @ the pun.. :biggrin:

after reading the posts i was quite confused about how to get the force when the bb wasn't in equilibrium, but i guess it triggered something because i got it the next time i tried.. :smile:

thanks to all once again.. :smile:

generalebriety, how can you tell that its retardation, is it because acceleration is 4/5g, so you have to multiply it with -9.8, which makes it negative? :s-smilie:

Retardation is just acceleration "the other way". You know it's acceleration down the plane, because the resultant force is down the plane. Simple application of Newton's second law. :smile:

"g = -9.8" is a misunderstanding on your part, I think. Maybe you're thinking of projectiles or something, where it's often simpler to choose g = -9.8 because the particle is initially moving upwards, we often want to measure the distance upwards (you rarely get projectiles fired downwards :biggrin:

), but the acceleration is awkward and decides to go down. If it helps you, g = -9.8j... but that doesn't mean you must always choose g to be negative. It's often more convenient to choose "down" as the "positive" direction, if you're trying to measure a velocity that something is falling at, or trying to measure the distance down from a certain point, or something. :smile:

n0b0dy

ah, the question gets annoying as I move on..

b) Given that the machine is placed at a point A, 30m from the top edge of the plane, and the ball-bearings are projected with an initial speed of 20ms-1,
find, giving your answer to the nearest cm, how close the bb get to the top edge of the plane?

do we use equations of motion, u=20, a= 7.84, s=?, t=?, v=?

great, i only know acceleration and initial speed, how do i get the max distance the bb travels?

You know the final speed too. It stops moving when it gets to the highest point: v=0.

Reply 16

OP

generalebriety

Retardation is just acceleration "the other way". You know it's acceleration down the plane, because the resultant force is down the plane. Simple application of Newton's second law. :smile:

"g = -9.8" is a misunderstanding on your part, I think. Maybe you're thinking of projectiles or something, where it's often simpler to choose g = -9.8 because the particle is initially moving upwards, we often want to measure the distance upwards (you rarely get projectiles fired downwards :biggrin:

), but the acceleration is awkward and decides to go down. If it helps you, g = -9.8j... but that doesn't mean you must always choose g to be negative. It's often more convenient to choose "down" as the "positive" direction, if you're trying to measure a velocity that something is falling at, or trying to measure the distance down from a certain point, or something. :smile:

You know the final speed too. It stops moving when it gets to the highest point: v=0.

yay!! done it! :biggrin:

thank you muchas.. :p:

Reply 17

No problem. :smile:

Reply 18

OP

just to make sure, could anyone confirm my answer to the part c of the question?

c) how long does it take for a bb to travel from the highest point it reaches back down to the point A again?

here's what i di:

u= 0
a= 7.84
s= 30
t = ?

s = ut + 1/2at[sq]
30= 0t + 1/2 * 7.84 * t[sq]
30 = 3.92t[sq]
t[sq] = 7.65
t = 2.77 secs (2d.p.)

thank you.. :smile:

Reply 19