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TrueTears · « **Reply #165 on:** January 10, 2009, 04:51:35 pm »

yeah thanks i get all of that now

TrueTears · « **Reply #166 on:** January 10, 2009, 07:49:01 pm »

Also this question

1. In a lift that is accelerating upwards at 2m/s^2 a spring balance shows the apparent weight of the body to be 2.6 kg wt. What would be the reading if the lift were at rest?

(and also btw what does kg wt mean?)

Mao · « **Reply #167 on:** January 10, 2009, 09:57:15 pm »

1 kg wt is '1 kilogram weight', i.e. 1g N

When accelerating upwards, the lift exerts a force upwards on the weight, i.e. the weight exerts an opposite force down towards the lift, in the same direction as gravity. Hence, the normal reaction force would be
$(2+g)\times m=2.6g\; kg\; wt\implies m\approx \frac{2.6\times 9.8}{11.8}\approx 0.216 \; kg$

hence when not accelerating, the mass would exert 0.216 kg wt of force on the spring balance.

TrueTears · « **Reply #168 on:** January 11, 2009, 12:48:48 am »

thanks mao

TrueTears · « **Reply #169 on:** January 11, 2009, 12:50:17 am »

and also just a tiny bit stuck on Q 12, i think my 2nd equation is missing something...

Mao · « **Reply #170 on:** January 11, 2009, 12:20:52 pm »

8kg mass: $T_2=8a$

12kg mass: $T_1-T_2 = 12a$

5kg mass: $5g - T_1 = 5a$

TrueTears · « **Reply #171 on:** January 11, 2009, 04:41:57 pm »

thanks mao

also:
1. A body of mass m is projected vertically upwards with speed u. Air resistance is equal to k times the square of the speed where k is a constant. find the maximum height reached and the speed when next at the point of projection

2. Two equal forces of 10 N act on a particle. The angle between the 2 forces is 50 deg. Find the magnitude and direction of the single force which when applied will hold the particle in equilibrium.

So for Q 2 after some working the magnitude resultant force of the two 10 N forces is 18.126N. So we need 18.126 N of force to balance it. For equilibrium both the j and i components must be equal to 0. So resolving the j component we have $10 + 10cos50 +18.126cos(\alpha) = 0$

and resolving the i component we have $10 cos(40) + 18.126cos(\beta) = 0$

But solving for $\alpha$ for the j component has 2 answers and solving for $\beta$ for the i component also has 2 answers, which answer is right? And is this the right way to do this question?

Mao · « **Reply #172 on:** January 11, 2009, 06:36:03 pm »

1.
where is this question from? It's rather damn complicated if there's only pronumerals to deal with.

2.
not sure why you are doing it this way. Since the two forces are equal and acting at 50 degrees from each other, you should define i as the 25^o direction, hence the j component of the two forces cancel out. You will just have an 18.126N force acting at 180-25=155 degrees from either force.

TrueTears · « **Reply #173 on:** January 11, 2009, 06:49:20 pm »

1. yeah, its from the spesh essential book, only pronumerals no numbers., ive tried it for hours , the equations are annoying lol

2. o i see, stupid me ><

/0 · « **Reply #174 on:** January 11, 2009, 08:55:00 pm »

I love equations likes this... part of the reason I like physics so much

Let up be Positive and down be Negative for this question.

$F_{net}=F_g+F_{air} =-mg- kv^2 \implies a_{net} = -g-\frac{kv^2}{m}$

$v\frac{dv}{dx} = -(g+\frac{kv^2}{m})$

$\frac{dv}{dx} = \frac{-(mg+kv^2)}{mv}$

$\frac{dx}{dv} = \frac{-mv}{mg+kv^2}$

$x=\frac{-m}{2k}\log_e|kv^2+mg|+C$

When $x=0$ , $v=u$ :

$\therefore C = \frac{m}{2k} \log_e|ku^2+mg|$

$\therefore x=\frac{-m}{2k}\log_e|kv^2+mg|+\frac{m}{2k}\log_e|ku^2+mg|=\frac{m}{2k}\log_e\left|\frac{ku^2+mg}{kv^2+mg}\right|$

For max height:

When $v=0$ ,

$x_H=\frac{m}{2k}\log_e\left|\frac{ku^2+mg}{mg}}\right|=\frac{m}{2k}\log_e\left|1+\frac{ku^2}{mg}\right|$

And since none of the pronumerals are 'likely' to be negative you might get away with round brackets...

For part b), we have $a_{net} = -g+\frac{kv^2}{m}$

$\frac{dv}{dx} = \frac{kv^2-mg}{mv}$

$\frac{dx}{dv}=\frac{mv}{kv^2-mg}$

$x = \frac{m}{2k}\log_e|kv^2-mg|+C$

When $x = x_H$ , $v=0$

$\implies C = \frac{m}{2k}\log_e\left|\frac{mg+ku^2}{m^2g^2}\right|$

$\therefore x = \frac{m}{2k}\log_e\frac{|(kv^2-mg)(ku^2+mg)|}{m^2g^2}$

When $x=0$ ,

$|(kv^2-mg)(ku^2+mg)| = m^2g^2$

$(kv^2-mg)(ku^2+mg) = \pm m^2g^2$

But $kv^2<mg \implies (kv^2-mg)(ku^2+mg) = -m^2g^2$

$\therefore v = u\sqrt{\frac{mg}{ku^2+mg}}$

TrueTears · « **Reply #175 on:** January 11, 2009, 09:33:34 pm »

legendary ST

TrueTears · « **Reply #176 on:** February 07, 2009, 05:58:49 pm »

yay spesh is finally back ! gonna repost some more Q in this thread nows XD

lacoste · « **Reply #177 on:** February 07, 2009, 06:59:50 pm »

Quote from: /0 on January 11, 2009, 08:55:00 pm

I love equations likes this... part of the reason I like physics so much

But $kv^2<mg \implies (kv^2-mg)(ku^2+mg) = -m^2g^2$

$\therefore v = u\sqrt{\frac{mg}{ku^2+mg}}$

Damn, spesh looks hard, damn, man, and you love these q's

TrueTears · « **Reply #178 on:** February 07, 2009, 07:57:49 pm »

Trivial q... for ellipses do we have to put the co-ordinates of the X and Y extremities?

Mao · « **Reply #179 on:** February 07, 2009, 08:07:50 pm »

I would.

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