Force and work.

- Babynini

Force and work.

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- Babynini

A car drives at 800ft. on a road inclined at 9 degrees. Car weighs 2700lb. Thus gravity acts straight down on the car with a constant force of f=-2700j. Find the work done by the car in overcoming gravity.

- Babynini

Um..what i've done so far is
|v|=2700cos(9) = 2666.76
which is the force experienced by the driveway. Is that the answer i'm looking for?

- Babynini

@ganeshie8

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## More answers

- Babynini

@zepdrix :)

- Babynini

@jim_thompson5910

- jim_thompson5910

Work = Force * Displacement

- jim_thompson5910

Gravity is pulling straight down, so when we think of "displacement" we are only considering the vertical displacement of the car

- jim_thompson5910

|dw:1433399821064:dw|

- jim_thompson5910

when you wrote `Thus gravity acts straight down on the car with a constant force of f=-2700j`
that doesn't make any sense. Joules aren't a measure of force. They are measure of energy. You might be thinking of newtons?

- Babynini

uhh idk that's what my homework says.

- jim_thompson5910

oh nvm, -2700j is a vector, not 2700 joules

- Babynini

haha yeah

- Babynini

where is hte 800 from?

- jim_thompson5910

"A car drives at 800ft"

- jim_thompson5910

I'm assuming it drives up that incline and drives 800 ft

- Babynini

ah yeah k.

- jim_thompson5910

use trig to find the value of y in the drawing below
|dw:1433400113262:dw|

- Babynini

125.148 ?

- jim_thompson5910

so the car essentially goes up vertically 125.148 ft
this is the vertical displacement we care about

- Babynini

(ps: it says you are expected to use vectors and the appropriate formulas to find work. Just as a note. :P)

- Babynini

oo ok

- Babynini

so now we have
2=-2700 * 125.148

- Babynini

that 2 was meant to be a w

- Babynini

er. that will make a very nasty negative number..

- jim_thompson5910

were you given the vector formula for work in physics?

- Babynini

This is precalc.

- Babynini

give me a moment to find it, I have a formula somewhere :)

- Babynini

Yep
w=f*d

- jim_thompson5910

I'm thinking you basically take the force needed to climb that vertical distance and multiply it with the vertical distance
I'm not sure how to fit in vector formulas though

- Babynini

so an example given is:
A man pulls a wagon horizontally by exerting a force of 20 lb on the handle. If the handle makes an angle of 60 degrees with horizontal, find the work done in moving the wagon 100ft
Solution:
We choose a coordinate system with the origin at the initial position of the wagon. That is, the wagon moves from point p(0,0) to the point Q(100,0) the vector that represents this displacement is
d=100i
The force on the handle can be writen in terms of components as
F= (20cos(60))i + (20sin(60))j
F = 10i+10 sq3j
Thus the work done is
(10i+10 sq3j)*(100i)=1000
= 1000

- Babynini

so our displacement is
D = 800i

- Babynini

so it's written as
F=(2700cos9)i+(2700sin9)j
= 2666.785i+422.373j

- Babynini

and work done is =
(2666.785i+422.373j)*(800i)

- Babynini

I'm not sure how they calculated the last part though.

- Babynini

Is this kind of making sense? I'm following formulas mostly. o.0

- Babynini

@jim_thompson5910 :P

- Babynini

(sorry, i'll be patient ;))

- jim_thompson5910

sorry I got distracted for a sec
but it looks like you're on the right track. You dot product the force vector F and the displacement vector d to get the work

- Babynini

it's ok! I know you're helping other people. Keep up the hard work :)

- Babynini

so is
2133829.189 the final answer you're getting?

- jim_thompson5910

The force of gravity is
F = 0i + (-2700j)
essentially this vector pulls the car straight down with magnitude 2700 lb
--------------------
The displacement vector d = 800*cos(9)i + 800*sin(9)*j
--------------------
dot product the two vectors
W = F dot d
W = [0i + (-2700j)] dot [800*cos(9)i + 800*sin(9)*j]
W = [0*800*cos(9)] + [-2700*800*sin(9)]
W = -337,898.444486899
That seems like a pretty massive number, so I'm not 100% sure

- Babynini

bleh yeah that's pretty huge :/

- Babynini

wait why 800 instead of 2700 before the cos and sin?

- jim_thompson5910

because if we ignore the weight, and we just consider moving along the hypotenuse, we basically have this triangle
|dw:1433402120340:dw|

- Babynini

but in the example one they used the weight of the cart in that spot of the equation, which is why I ask.
hm but yeah I see what you're saying.

- jim_thompson5910

|dw:1433402181444:dw|

- jim_thompson5910

|dw:1433402210524:dw|

- jim_thompson5910

hmm now I'm not sure. I've looked back at your example and you're right

- Babynini

lol. What is this madness.

- anonymous

\[W=\vec F. \vec s=Fs \cos(\theta)\]

- anonymous

Everything is given, just convert to standard units and use formula, feet to metres, pounds to kilograms and degrees to radians

- anonymous

|dw:1433411117463:dw|

- Babynini

@rational how do I solve this? There's so many formulas that have been given here D:

- Babynini

F = -2700j so = <0,-2700>
Displacement (or shift) = 800 feet. so = <800,0>
all correct so far?

- jim_thompson5910

<800,0> means you go 800 ft to the right and 0 ft up/down

- jim_thompson5910

<800,0> = 800i + 0j

- Babynini

Yep yep.
So now according to W = F * D
I dot product
<0,-2700> * <800,0> ?

- jim_thompson5910

but the object isn't going on a flat level plane
its on an inclined plane

- Babynini

oh yeh.

- Babynini

where is that incorporated?

- jim_thompson5910

think of the point \(\Large (r, \theta) = (800, 9^{\circ})\)
what is the rectangular (x,y) form of this point?

- Babynini

what?

- jim_thompson5910

|dw:1433463727512:dw|
what are x and y?

- Babynini

Opposite and Adjacent? is that what you're asking?

- jim_thompson5910

Once you find x and y, you will be able to represent the displacement as a vector

- jim_thompson5910

you use trig to find x and y
sin(theta) = opp/hyp
sin(9) = y/800
y = 800*sin(9)
do the same with x (but use cosine instead)

- Babynini

y= 125.149
x=790.151

- jim_thompson5910

|dw:1433464222924:dw|

- jim_thompson5910

so the displacement vector is <790.151, 125.149> or 790.151i + 125.149j

- Babynini

and Force still is = <0,-2700> ?

- jim_thompson5910

I'm realizing now that's where the mistake is
that's the force of gravity, but we want the force of the car
if it asked "what is the work done by gravity?" then we would use F = <0,-2700>

- jim_thompson5910

but instead it says "Find the work done by the car in overcoming gravity."

- Babynini

aah hrm. so something to do with the weight?

- Babynini

Wait, but it says "the gracity acts straight down on the car with a constant force of -2700j." the force pushing against the car would be the same the car is pushing back though. sot hat would be the force the gar is "pushing against gravity"?

- Babynini

*gravity

- jim_thompson5910

maybe to overcome the force of -2700j, the car needs to apply 2700 lb somehow?

- jim_thompson5910

yeah as you just said

- jim_thompson5910

I'm not sure though

- Babynini

But -2700 is what is being pushed against the car. Which is the same amount the car is pushing back against gravity. So that should still be the force.

- jim_thompson5910

that might be correct

- Babynini

...

- jim_thompson5910

I found this page
http://cposcience.com/home/Portals/2/Media/post_sale_content/PHY2/Ancillaries/SkillSheets/Unit_4/10.2_WorkDoneagainst.pdf

- jim_thompson5910

sorry I'm not that great at physics

- Babynini

apparently neither am I ;P

- jim_thompson5910

if you use that link's formula, be careful to make sure things are converted to metric

- Babynini

er but this is vector stuff not joules and what not

- jim_thompson5910

pounds ---> kilograms
feet ----> meters

- jim_thompson5910

joules is just the units of work in physics

- Babynini

Ah k.

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