ATAR Notes: Forum

Uni Stuff => Science => Faculties => Physics => Topic started by: bucket on March 16, 2009, 07:11:44 pm

Title: Foundation(VCE) Physics Questions :S
Post by: bucket on March 16, 2009, 07:11:44 pm: Mmm I really did not know where to post this question, it's a university subject but at VCE level I assume.

"Vector $V_1$ is 8.08 units long and points along the negative x-axis, Vector $V_2$ is 4.51 units long and points at +45^o to the positve x-axis. Determine the sum of the two vectors."
Title: Re: Foundation(VCE) Physics Questions :S
Post by: TrueTears on March 16, 2009, 07:19:07 pm: $V_1 = -8.08i$

$V_2 = 3.189i + 3.189j$

now just add those 2 and this yields $-4.891i + 3.189j$

To work out $V_2$ basically just draw the vector and use trignometry.

$sin(45) = \frac{x}{4.51}$

x = 3.189 (3 dp)

$cos(45) = \frac{z}{4.51}$

z = 3.189(3 dp)
Title: Re: Foundation(VCE) Physics Questions :S
Post by: bucket on March 16, 2009, 07:21:08 pm: Apparently the answer is 5.84 and 33.1^o above -x-axis
Title: Re: Foundation(VCE) Physics Questions :S
Post by: TrueTears on March 16, 2009, 07:22:54 pm: Quote from: bucket on March 16, 2009, 07:21:08 pm
Apparently the answer is 5.84 and 33.1^o above -x-axis

yeap then just do this

let the vector sum be $Y$

so $Y = -4.891i+3.189j$

now $|Y| = \sqrt{(-4.891)^2 + (3.189)^2} = 5.84$

to work out the angle, again draw the vector Y then use trig

so, $tan(\theta) = \frac{3.189}{4.891}$

$\theta \approx 33.1^o$

both way are acceptable. I guess I should have answered in this format because the question state the vectors like that, so yeah |Y| means the magnitude of Y, ie the length in units, and the angle is just how far up it is from the x axis ( in this case it's from the negative x axis)
Title: Re: Foundation(VCE) Physics Questions :S
Post by: bucket on March 16, 2009, 07:25:35 pm: Perfect!
Cheers mate.
Man I feel so dumb for not being able to do this stuff :p

MAN ITS SO SIMPLE!! AHRHJKFHGJKFHGJKF lol
Title: Re: Foundation(VCE) Physics Questions :S
Post by: TrueTears on March 16, 2009, 08:30:57 pm: Quote from: bucket on March 16, 2009, 07:25:35 pm
Perfect!
Cheers mate.
Man I feel so dumb for not being able to do this stuff :p

MAN ITS SO SIMPLE!! AHRHJKFHGJKFHGJKF lol

lol I always make stupid mistakes, asif blame yourself over this question haha maybe you were just stomped :P
Title: Re: Foundation(VCE) Physics Questions :S
Post by: enwiabe on March 16, 2009, 09:29:30 pm: do you have ali moghimi
Title: Re: Foundation(VCE) Physics Questions :S
Post by: bucket on March 22, 2009, 02:38:39 pm: What is the maximum speed with which a 1050kg car can round a turn of radius 70m on a flat road if the coefficient of friction between tires and the road is 0.80? Is this result independent of the mass of the car?
Title: Re: Foundation(VCE) Physics Questions :S
Post by: TrueTears on March 22, 2009, 02:49:00 pm: Yes this is independent of the car's mass

$F_{Net} = \frac{mv^2}{r}$

$F_{Net} = F_{sideways friction} = \beta \times N$ (let $\beta$ = coefficient of friction = 0.80)

N is normal force = mg

equating yields $\beta \times N = \frac{mv^2}{r}$

But $N = mg$

$\beta \times mg = \frac{mv^2}{r}$

the mass cancels which leaves us with:

$\beta \times g = \frac{v^2}{r}$

to work out max speed, sub in the values and work out v :)
Title: Re: Foundation(VCE) Physics Questions :S
Post by: bucket on March 22, 2009, 02:53:40 pm: ahhh how do you guys learn how to approach these problems :S it's impossible for me to teach myself this !!
Title: Re: Foundation(VCE) Physics Questions :S
Post by: TrueTears on March 22, 2009, 02:57:07 pm: Just mess around with the equations :P

EDIT: what the... I am starting to see red boxes for LaTeX now lol
Title: Re: Foundation(VCE) Physics Questions :S
Post by: bucket on March 22, 2009, 02:58:31 pm: I get confused whenever I'm faced with a question involving friction :S. How do you work out friction required to round a curve? Does the force of the friction have to be the same as the centripetal force?? lol
Title: Re: Foundation(VCE) Physics Questions :S
Post by: TrueTears on March 22, 2009, 03:04:00 pm: Quote from: bucket on March 22, 2009, 02:58:31 pm
I get confused whenever I'm faced with a question involving friction :S. How do you work out friction required to round a curve? Does the force of the friction have to be the same as the centripetal force?? lol

Assuming constant circular motion. $F_{net} = F_{sideways friction}$
Title: Re: Foundation(VCE) Physics Questions :S
Post by: bucket on March 22, 2009, 03:08:22 pm: hmm, I see.

I have another question lol...
"A 1000kg sports car moving at 20m/s crosses the rounded top of a hill (r=100m). Determine the normal force on the car."

When I tried to do this question I assumed mg=F_N which was obviously wrong and stupid to do... but I have no idea how else to approach this!
The answer is 5800N
Title: Re: Foundation(VCE) Physics Questions :S
Post by: Flaming_Arrow on March 22, 2009, 03:11:35 pm: hint: use $\frac{mv^2}{r}- mg$
Title: Re: Foundation(VCE) Physics Questions :S
Post by: TrueTears on March 22, 2009, 03:12:43 pm: Let up be +ve and down be -ve

now let's analysis all the forces on the car

we got N going up, mg going down and $F_{net}$ going down. (if the net force was going up, then the car would not be on the ground! lol)

so $N - mg = -\frac{mv^2}{r}$

$N - 1000(9.8) = -\frac{1000(20)^2}{100}$

solving for N yields N = 5800N
Title: Re: Foundation(VCE) Physics Questions :S
Post by: bucket on March 22, 2009, 03:51:34 pm: Ah i see...but why does the sum of the forces equal the centripetal force? lol
Nevermind... so simple... because the centripetal force is the net force
duh.
sorry lol
Title: Re: Foundation(VCE) Physics Questions :S
Post by: TrueTears on March 22, 2009, 03:55:06 pm: Quote from: bucket on March 22, 2009, 03:51:34 pm
Ah i see...but why does the sum of the forces equal the centripetal force? lol
Nevermind... so simple... because the centripetal force is the net force
duh.
sorry lol
yeap, the sum of all the REAL forces = Net force
Title: Re: Foundation(VCE) Physics Questions :S
Post by: bucket on March 22, 2009, 04:17:36 pm: Another...lol

"A 1200kg car rounds a curve of radius 70m banked at an angle of 12^o. If the car is traveling at 90km/h, will a friction force be required? If so how much and in what direction?"
Title: Re: Foundation(VCE) Physics Questions :S
Post by: TrueTears on March 22, 2009, 04:26:15 pm: $F_{net} = F_{friction}cos(\theta) + Nsin(\theta)$

Now draw a diagram and split everything into its parallel and perpendicular components, this yields:

$Ncos(\theta) = mg$

$Ncos(12) = 1200 \times 9.8$

N = 12022.73

so $\frac{mv^2}{r} = F_{friction}cos(12) + 12022.73sin(12)$

$\frac{1200(25)^2}{70} = F_{friction}cos(12) + 12022.73sin(12) ( 90km/h = 25m/s)$

solving for $F_{friction}$ yields 8398.14N

direction: towards centre of the curve

EDIT: sorry forgot to square the $v^2$ -_-""
Title: Re: Foundation(VCE) Physics Questions :S
Post by: bucket on March 22, 2009, 04:39:00 pm: Thanks heaps man.