Post by: Ken on October 31, 2008, 09:23:35 pm
Post by: ed_saifa on October 31, 2008, 09:28:45 pm
Post by: Ken on October 31, 2008, 09:43:06 pm
Post by: ed_saifa on October 31, 2008, 09:44:09 pm
Post by: Ken on October 31, 2008, 09:46:13 pm
Post by: ed_saifa on October 31, 2008, 09:47:22 pm
Post by: Ken on October 31, 2008, 10:00:08 pm
Post by: ed_saifa on October 31, 2008, 10:08:06 pm
This explain the conjugate root theorem which makes this question a lot easier.
Post by: fredrick on October 31, 2008, 10:39:07 pm
Q)If one of the roots of
So
Now
Equate coefficents:
and
So
Now just quadratic equation or CTS and you get
there the right answers i hope
Post by: shinny on October 31, 2008, 11:11:41 pm
Post by: AppleXY on October 31, 2008, 11:24:31 pm
only takes a few minutes lool
Post by: fredrick on October 31, 2008, 11:31:29 pm
Post by: Ken on October 31, 2008, 11:54:32 pm
Post by: shinny on October 31, 2008, 11:59:58 pm
Post by: quangiez on November 01, 2008, 12:27:23 pm
It asks for the value of D if the box is 'not at the point of moving across the floor'. Should the answer be R \ {0.5g}?
Post by: shinny on November 01, 2008, 12:32:14 pm
Post by: quangiez on November 01, 2008, 12:38:11 pm
'at the point of moving', D = 0.5g
and because they ask you to find the value of D when it's not 'at the point of moving' you exclude 0.5g and include all other values. Or maybe im just making a fool out of myself -.-
Post by: shinny on November 01, 2008, 12:42:33 pm
Post by: dcc on November 01, 2008, 12:46:31 pm
But the question states the box is at rest also, so D can't be above that value either. Therefore it'd be D<0.5g i.e.(as is in the solutions), or even more correct possibly might be
A stronger inequality would be
Post by: polky on November 01, 2008, 01:00:54 pm
Post by: shinny on November 01, 2008, 01:02:29 pm
Post by: Mao on November 01, 2008, 01:27:56 pm
But the question states the box is at rest also, so D can't be above that value either. Therefore it'd be D<0.5g i.e.
A stronger inequality would be, as the box may be moving in the other direction. A question similar to this has appeared in a VCAA exam, and (lots of) students forgot about the possibility of a force in the opposite direction.
I am under the impression that D in this question refers to the magnitude.
Post by: dcc on November 01, 2008, 01:29:51 pm
How could it move in the other direction o_O He's only PULLING one way, and you define where D is yourself; whether you say he's pulling left or right doesn't matter as it just horizontally flips the force diagram. You can't produce a negative force in the direction in which you are pulling by pulling something...
Are you saying if you push it in the opposite direction, it will never move? Because you can't say that a positive force denotes a direction to both the left and the right.
If you provided a force of
Simiarly, if you provided a force of
Clearly the force must lie between these two bounds, otherwise, taking your argument to its logical conclusion, you are suggesting that
But the question states the box is at rest also, so D can't be above that value either. Therefore it'd be D<0.5g i.e.
A stronger inequality would be
I am under the impression that D in this question refers to the magnitude.
If it is the magnitude, then the answer
Post by: Mao on November 01, 2008, 01:46:33 pm
A box of mass 5kg rests on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.1. A boy applies a horizontal dragging force of D newtons to the box in an attempt to move it.
Find values of D if the box is not at the point of moving across the floor.
now, D newtons suggests to me that D is magnitude.
the question also did not specify a direction [nor any sense of it, other than force is parallel to the direction of motion], and that it was a pulling/dragging force. I believe shinjitsuzx is somewhat correct in saying that since the kid is pulling it in one way, we would naturally assume that is the positive direction. Thus he cannot exert a negative force as that would be "pushing", not pulling. Though a negative force could potentially move it, it is not plausible to assume that the kid can actually exert a negative force on the box.
had the question been worded "A boy applies a horizontal force", we may argue that it could both be pull and push, though that can itself be dismissed by stating "let the direction of force exerted by the boy on the box be the positive direction"
Post by: dcc on November 01, 2008, 01:57:23 pm
now, D newtons suggests to me that D is magnitude.
That suggests to me that they would like the answer in Newtons, more then anything. VCAA sometimes represents forces using vector notation, and even then, they do not neglect to include the fact the force is in Newtons. If they meant magnitude, they would of said 'a dragging force of magnitude D newtons'.
the question also did not specify a direction [nor any sense of it, other than force is parallel to the direction of motion], and that it was a pulling/dragging force. I believe shinjitsuzx is somewhat correct in saying that since the kid is pulling it in one way, we would naturally assume that is the positive direction. Thus he cannot exert a negative force as that would be "pushing", not pulling. Though a negative force could potentially move it, it is not plausible to assume that the kid can actually exert a negative force on the box.But what if this assumption is not made? What if I were to define positive to the right, and assume the boy was applying a force to the left. I would have an entirely different set of inequalities, which would be just as correct as the ones provided above.
The safest, most correct way to answer this question would be to make no assumptions about the direction of the force, and care only about the magnitude of said force. That way, there are no problems with the plausibility of a "negative pushing force", as all directions are rendered equal.
Post by: shinny on November 01, 2008, 02:17:30 pm
But what if this assumption is not made? What if I were to define positive to the right, and assume the boy was applying a force to the left. I would have an entirely different set of inequalities, which would be just as correct as the ones provided above.
The safest, most correct way to answer this question would be to make no assumptions about the direction of the force, and care only about the magnitude of said force. That way, there are no problems with the plausibility of a "negative pushing force", as all directions are rendered equal.
Ah right so that's what you meant. I thought you were talking about fixing the i-j system, and changing the direction of the force itself (which wouldn't be possible as he's pulling). If you mean in terms of changing the i-j directions, then yeh I guess that is true, but I doubt VCAA tends to care over technicalities such as that since like Mao said, people just tend to define the direction of movement as the positive direction.