#1

I need some help one specific vector problem:

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150cm and makes an angle of 120 degrees with the positive x-axis. The resultant displacement has a magnitude of 140cm and is directed at an angle of 35 degrees to the positive x-axis. Find the magnitude of the second displacement.

Me and a friend tried figuring it out but what I came up with was nothing like what was in the back of the book. Some help please? Thanks a million

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150cm and makes an angle of 120 degrees with the positive x-axis. The resultant displacement has a magnitude of 140cm and is directed at an angle of 35 degrees to the positive x-axis. Find the magnitude of the second displacement.

Me and a friend tried figuring it out but what I came up with was nothing like what was in the back of the book. Some help please? Thanks a million

#2

Work out the vertical and horizontal components of the first displacement and the resultant displacements and you can easily find the second displacement from those.

#3

break it down into vector components and add them from there

#4

I did. And then I added them but what I got was much more than the answer in the back of the book.

150sin(120)

150cos(120)

and

140sin(35)

140cos(35)

Right? then add them up

150sin(120)

150cos(120)

and

140sin(35)

140cos(35)

Right? then add them up

#5

you could try using trig to solve this

#6

is your calculator set in degrees?

if its in radians it will give you the wrong sin/cos numbers.

if its in radians it will give you the wrong sin/cos numbers.

#7

It is in degree.

After i get my components I square the x and y and then take the square root (phythagorean) right?

Gah...what am i doing wrong?

After i get my components I square the x and y and then take the square root (phythagorean) right?

Gah...what am i doing wrong?

#8

Yeah...that gives you the component lengths so use pythagoras to find the magnitude.I did. And then I added them but what I got was much more than the answer in the back of the book.

150sin(120)

150cos(120)

and

140sin(35)

140cos(35)

Right? then add them up

Edit: ^ not sure

#9

I got about 210 but the book says 196cm at 345 degrees though i don't know how i'm supposed to get that.

#10

This is how my teacher taught it to me: set up a box that sets up x and y components of each vector. Than, its simple addition/subtraction to figure out what components vector two needs to make the resultant vector.

Sorry bout the bad writing

Edit: So x-component is 189.7cm, and y-component is -49.9. ((189.7^2)+(-49.9^2))^.5 equals 196.15cm as the magnitude. cos(theta)=(189.7/196.15), and you get 14.73 degrees. (about 360-345)

Sorry bout the bad writing

Edit: So x-component is 189.7cm, and y-component is -49.9. ((189.7^2)+(-49.9^2))^.5 equals 196.15cm as the magnitude. cos(theta)=(189.7/196.15), and you get 14.73 degrees. (about 360-345)

*Last edited by orcs4life at Sep 11, 2008,*

#11

I got those values -_-

#12

read my edit

Edit: I dont know why the book says 345 degrees, based on the values in the problem that is impossible. Maybe -345 degrees?

Edit: I dont know why the book says 345 degrees, based on the values in the problem that is impossible. Maybe -345 degrees?

#13

It's an algebra problem

V1 + V2 = Vresult

You know V1 and Vresult, so

V2 = Vresult - V1

So

V1 + V2 = Vresult

You know V1 and Vresult, so

V2 = Vresult - V1

So

*subtract*(not add) the x and y components for V1 from the x and y components for Vresult and that's your answer.
#14

I GOT IT!!!! (lol edit)

Thanks SO much. you subtact the y components giving you a POSITIVE not NEGATIVE. then take the tan^-1(y/x)

Then add 360

Again, thanks.

Thanks SO much. you subtact the y components giving you a POSITIVE not NEGATIVE. then take the tan^-1(y/x)

Then add 360

Again, thanks.

*Last edited by marktuazon at Sep 11, 2008,*

#15

Funny, I...uh...have a college aged storm trooper who...uh...was doing that same exact problem the other night...yeah...

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