Post by: yabbaboo on November 28, 2011, 07:49:30 pm
I got the answer but it is wrong. The answer is 2.46 m. Can anyone help me please?
Post by: b^3 on November 28, 2011, 08:06:45 pm
That is
On the 28 kg mass
28g N down
T in the rope up (tension)
On the 68 kg mass
T in the rope from the 68 kg mass
68g N vertically down on the 68 kg mass
and R (reaction force) perpendicular to the surface
Now next you need two simulataneous equations.
So on the 28 kg mass
T-28g=ma
T-28g=28a....1
On the 68 kg mass - Take down the slope as positive.
Resolving peroendicular to the slope -
R=68g*cos(20)
also
Resolving paralell to the slope - 68g*sin(20)-T=68a....2
From 1
T=28a+28g
sub into 2
68gsin(20)-28a-28g=68a
g(68sin(20)-28)=96a
g=9.8
a=-0.4814 m/s^2
So that is up the slope. (since down the slop was taken as positive and the acceleration turned out to be negative, so it is accelerating in the opposite direction).
now using our equ of motions
a=0.4814 m/s^2
u=0m/s
x=?
t=3.2s
x=ut+1/2*at^2
x=(0*3.2)+1/2 *(0.4814)*(3.2)^2
x=2.479m
x=2.48m
So it travels up the plane and after 3.2 seconds it has travelled 2.48m.
The answer is slightly out due to rounding somewhere, I'll see if I can find it. EDIT 2: Can't seem to fix it, maybe the answer is slightly wring, unless I've made a mistake.
EDIT: added image.
Post by: Bhootnike on November 28, 2011, 09:41:08 pm
Draw the forces on the diagram that is acting.
That is
On the 28 kg mass
28g N down
T in the rope up (tension)
On the 68 kg mass
T in the rope from the 68 kg mass
68g N vertically down on the 68 kg mass
and R (reaction force) perpendicular to the surface
Now next you need two simulataneous equations.
So on the 28 kg mass
T-28g=ma
T-28g=28a....1
On the 68 kg mass - Take down the slope as positive.
Resolving peroendicular to the slope -R-68gcos(20)=0- Not needed here since no friction
R=68g*cos(20)
also
Resolving paralell to the slope - 68g*sin(20)-T=68a....2
From 1
T=28a+28g
sub into 2
68gsin(20)-28a-28g=68a
g(68sin(20)-28)=96a
g=9.8
a=-0.4814 m/s^2
So that is up the slope. (since down the slop was taken as positive and the acceleration turned out to be negative, so it is accelerating in the opposite direction).
now using our equ of motions
a=0.4814 m/s^2
u=0m/s
x=?
t=3.2s
x=ut+1/2*at^2
x=(0*3.2)+1/2 *(0.4814)*(3.2)^2
x=2.479m
x=2.48m
So it travels up the plane and after 3.2 seconds it has travelled 2.48m.
The answer is slightly out due to rounding somewhere, I'll see if I can find it. EDIT 2: Can't seem to fix it, maybe the answer is slightly wring, unless I've made a mistake.
EDIT: added image.
I'm probably wrong, but shouldn't the weight force acting down on the weights be w= mg? So if g = 10, ( or 9.8), w= 280. And 680. N. Respectively?
Post by: Lasercookie on November 28, 2011, 09:48:49 pm
I'm probably wrong, but shouldn't the weight force acting down on the weights be w= mg? So if g = 10, ( or 9.8), w= 280. And 680. N. Respectively?
Draw the forces on the diagram that is acting.
That is
On the 28 kg mass
28g N down
T in the rope up (tension)
On the 68 kg mass
T in the rope from the 68 kg mass
68g N vertically down on the 68 kg mass
28g N = 28 * g N= 28 * 10 N = 280 N
He has considered w=mg, only that he didn't substitute in the value for g, instead leaving it as 28g and 68g.
Post by: b^3 on November 28, 2011, 09:49:55 pm
It is, just instead of using 280 and 680, I keep it as 28g and 68g and subbed g back in at the end.Draw the forces on the diagram that is acting.
That is
On the 28 kg mass
28g N down
T in the rope up (tension)
On the 68 kg mass
T in the rope from the 68 kg mass
68g N vertically down on the 68 kg mass
and R (reaction force) perpendicular to the surface
Now next you need two simulataneous equations.
So on the 28 kg mass
T-28g=ma
T-28g=28a....1
On the 68 kg mass - Take down the slope as positive.
Resolving peroendicular to the slope -R-68gcos(20)=0- Not needed here since no friction
R=68g*cos(20)
also
Resolving paralell to the slope - 68g*sin(20)-T=68a....2
From 1
T=28a+28g
sub into 2
68gsin(20)-28a-28g=68a
g(68sin(20)-28)=96a
g=9.8
a=-0.4814 m/s^2
So that is up the slope. (since down the slop was taken as positive and the acceleration turned out to be negative, so it is accelerating in the opposite direction).
now using our equ of motions
a=0.4814 m/s^2
u=0m/s
x=?
t=3.2s
x=ut+1/2*at^2
x=(0*3.2)+1/2 *(0.4814)*(3.2)^2
x=2.479m
x=2.48m
So it travels up the plane and after 3.2 seconds it has travelled 2.48m.
The answer is slightly out due to rounding somewhere, I'll see if I can find it. EDIT 2: Can't seem to fix it, maybe the answer is slightly wring, unless I've made a mistake.
EDIT: added image.
I'm probably wrong, but shouldn't the weight force acting down on the weights be w= mg? So if g = 10, ( or 9.8), w= 280. And 680. N. Respectively?
EDIT: yes as laseredd has said.
Post by: Bhootnike on November 28, 2011, 09:52:10 pm
Maybe cause you used 9.8 it came out a bit off the answer?
Post by: b^3 on November 28, 2011, 09:53:13 pm
Oh ok, well there you go :)I tried g=10 m/s^2 as well and it was off by more.
Maybe cause you used 9.8 it came out a bit off the answer?
Post by: dc302 on November 28, 2011, 09:57:16 pm
Post by: b^3 on November 28, 2011, 10:00:05 pm
You should always use g=9.8 in spesh.Yeh make sure you always check the front of the section, where it has the intructions. It will tell you what value of g to use. For spesh g=9.8m/s^2 (as dc302 said) and for physics its g=10 m/s^2. But always check just in case.
Post by: Lasercookie on November 28, 2011, 10:01:37 pm
That's my guess to whatever the source of the question has done. I would think that's excessive rounding off though.
Post by: b^3 on November 28, 2011, 10:02:37 pm
If you round off a to 0.48, you getGood spot laseredd.
That's my guess to whatever the source of the question has done. I would think that's excessive rounding off though.
Post by: dc302 on November 28, 2011, 10:09:24 pm
Post by: b^3 on November 28, 2011, 10:11:18 pm
Lol I know I'm just being pedantic but g is (at least in spesh) used as the number, not the acceleration. Ie, g=9.8, not 9.8 m/s^2Yes thats right, and being pedantic pays off in the end, small things make the difference between a 50 and a 49.
EDIT: Just note g=9.8 while the acceleration due to gravity = 9.8m/s^2
Post by: Lasercookie on November 28, 2011, 10:12:03 pm
Lol I know I'm just being pedantic but g is (at least in spesh) used as the number, not the acceleration. Ie, g=9.8, not 9.8 m/s^2Nothing wrong with being pedantic :P
But what is the reasoning behind not writing m/s^2?
Post by: b^3 on November 28, 2011, 10:14:57 pm
Quote
Take the acceleration due to gravity to have magnitude g m/s2, where g=9.8.So yeh as said in previous post.
Post by: yabbaboo on November 28, 2011, 10:31:16 pm
Post by: b^3 on November 28, 2011, 10:34:53 pm
thankyou for your help!! Just one more question (sorry if it sounds stupid!), why T-28g?Since I initially took down the slope as positive, that means that for the 28kg mass, the positive direction would be up. At first we don't know which way it will move, so I set up the two equations with the positive direction as above. So it looked as if the block was moving down the slope, so T would be greater than 28g since it was mving up, so T-28g. But we end up with a -ve acceleration so it is moving in the opposite direction i.e. up the slope.
EDIT: make sure that you have the blocks set up so that the same direction is positive, that way the equations will be set up the right way.
Post by: dc302 on November 28, 2011, 11:43:25 pm
Lol I know I'm just being pedantic but g is (at least in spesh) used as the number, not the acceleration. Ie, g=9.8, not 9.8 m/s^2Nothing wrong with being pedantic :P
But what is the reasoning behind not writing m/s^2?
It's also simpler to use letters for numbers by themselves, rather than with units. For example if you wanted to have a speed at 9.8m/s, and g = 9.8 m/s^2, then your speed is like 9.8s (which looks weird). But yeah, as b^3 said, VCAA uses this as a convention anyway.