Finding the a+ib form of a complex number

[charmedbeauty]

Homework Statement

given |z|=3, Arg(z)=5Pi/6

find a+ib form of the complex number.

Homework Equations

The Attempt at a Solution

so from the arg(z) we can say it lies in the second quad.

Since 5Pi/6 is equivalent to then 180-150 =30
so,
3(sqrt3/2 +1/2i)

but they had 3/2(-sqrt3+i)

why is it negative? since 150 is in the sec quad.

 

[BruceW]

Difficult to explain with words. the complex plane is usually written with the real number line on the x-axis (horizontal) and the pure imaginary number line on the y-axis (vertical). And the angle is measured anticlockwise from the positive side of the real number line. Somewhere, I think you are imagining it slightly differently. Which part of this is different to how you were thinking of it?

 

[tiny-tim]

hi charmedbeauty

(have a square-root: √ and a pi: π )

why is it negative? since 150 is in the sec quad.

cos150° is negative

(and x in the 2nd quadrant is also obviously negative)

 

[HallsofIvy]

The only question I have is why you are asking this question! Yes, your number is in the second quadrant. And any point in the second quadrant has negative "x" (real part for complex numbers).

 

[charmedbeauty]

hi charmedbeauty

(have a square-root: √ and a pi: π )cos150° is negative

(and x in the 2nd quadrant is also obviously negative)

Ok thanks, I was just making a silly mistake and not realising that the real part was negative, thanks for clearing it up guys.

On another note I know this is probably not the right place to post but, my question is...
in physics I am told that the static fric is much larger then the kinetic fric in most cases depending on the surface.

So I have been wondering how this applies to the big coal trains that are connected to say loads of about 100 coal carts. How is it that the lead cart (the one doing all the work) is able to pull the other 100 carts along ie how is the lead cart able to overcome the static friction of 100 loaded carts?? because when I picture it in my head I can only imagine the wheels slipping.

 

[BruceW]

The trick is that they are all on wheels. Start by imagining a loaded cart being pulled along by a rope, will friction between wheels and the ground actually resist the motion of the cart?

 

[charmedbeauty]

The trick is that they are all on wheels. Start by imagining a loaded cart being pulled along by a rope, will friction between wheels and the ground actually resist the motion of the cart?

I would think that it would oppose it but depending on the total amount of force being applied you should be able to get it moving.
But where I still have trouble understanding is that it is only the lead cart which the applied force is acting on. how does it get the force required to overcome the stat fric of the other 100 carts, because I imagine it wouldn't be a simultaneous movement between carts ie the first cart would move before the second and so on. so the first overcomes its own stat friction and it only has kinetic friction until the force of the other carts is 'felt'. But at this moment is where I have trouble grasping what's going on, because the only friction acting against the first cart is its own kinetic friction, but surely its force couldn't be enough to overcome the stat friction of the other 100 carts?
Otherwise its acceleration would be huge based on Ftot=ma
what am I not understanding correctly here?

 

[BruceW]

I would think that it would oppose it but depending on the total amount of force being applied you should be able to get it moving.

Nope, friction between the wheels and the ground does not cause opposition to motion. This is why they use carts on wheels instead of dragging them along the ground. (Although there is still 'rolling friction', axle friction and air resistance, but these are often much less than the friction which would be caused by dragging it across the ground).

To understand this, think of what happens when the cart begins moving. There is an initial frictional force on the wheel, but what effect does this have?