Magnitude of Force within incline?

[anglum]

Magnitude of Force within incline?

A block with mass of 5.5kg is held in equilibrium on a frcitionless incline of 34 degrees by the horizontal force F. The acceleration of gravity is 9.81m/s^2

What is the magnitude of F? answer in units of N

What is the magnitude of the normal force? answer in units of N


I converted 5.5 kg to 53.9N

then i used sin of 56 degrees * 53.9 to get the horizontal force of 44.68N

then i used cos of 56 degrees * 53.9 to get the normal force of 30.14N

however both answers were incorrect

 

[silvashadow]

because you need to use sin of 34 not 56 degrees. you have the basic concept of weight the the triangle it forms. but that triangle is proportional the incline, so they have the same angles.

 

[anglum]

ok but sin of 34 * 53.9 = 30.14N

and cos of 34 * 53.9 = 44.68N

so i get the same answers ... and they are wrong... that's why i am stuck

 

[johnsonandrew]

because you need to use sin of 34 not 56 degrees. you have the basic concept of weight the the triangle it forms. but that triangle is proportional the incline, so they have the same angles.

Yeah, use 34 degrees, not 56. Fh= Fg*(sin x)
Your angle of incline will be the angle you use in the computation.

so Fh= (54.0)*(sin 34) = 30.2 N
and Fn= (54.0)*(cos 34) = 44.8 N

Double check my work.. but I think that's how you do it..

 

[johnsonandrew]

Is there additional information in the problem you didn't include? Because I believe that's correct...

 

[anglum]

those answers are incorrect... that's why i am so lost

 

[anglum]

the drawing shows the incline... the slope of the incline is 34 degrees... the block on it is 5.5 kg which is 53.9N

it slopes up from left to right... there is a line horizontal that is labeled as F.. and above it is a arrow pointing down sayin 34 degrees as well... where this horizontal line F touches the block is in the top left corner of the block... then an angled line down off of the top left corner of the block slopes down the incline as well

 

[silvashadow]

those look correct to me too

 

[anglum]

i wish there was a way to draw out what they give u ... i tried to describe it the best i could

 

[anglum]

this is the best that i can describe the drawing shows the incline... the slope of the incline is 34 degrees... the block on it is 5.5 kg which is 53.9N

it slopes up from left to right... there is a line horizontal that is labeled as F.. and above it is a arrow pointing down sayin 34 degrees as well...where this horizontal line F touches the block is in the top left corner of the block... then an angled line down off of the top left corner of the block slopes down the incline as well

its an online class so when i enter the answer it tells me its wrong

 

[silvashadow]

did you round before the answer? maybe you need to round only after finding the answer. that can change it by .1 or .01 or whatever

 

[johnsonandrew]

and the horizontal comes from the left?

 

[anglum]

i tried answers ranging from 30..05 to 30.2 and the same for 44.65 to 44.70

and they were wrong

 

[johnsonandrew]

did you round before the answer? maybe you need to round only after finding the answer. that can change it by .1 or .01 or whatever

Yeah, since its an online course where you enter the answer that might throw it off. Try my answers, I was careful with significant figures...

 

[anglum]

i am uploading the picture of the problem online.. i will have the link for u to see it in one minute

 

[johnsonandrew]

But I believe since the horizontal force touches the top left corner, and not the center of the block, it does not equal F parallel, so the formula does not apply... Let me give it another shot on paper...

 

[anglum]

http://i199.photobucket.com/albums/aa314/anglum/physicshelp.jpg

 

[johnsonandrew]

i am uploading the picture of the problem online.. i will have the link for u to see it in one minute

Awesome, that would help a lot.

 

[anglum]

check that link for the drawing of the problem

 

[johnsonandrew]

Ok, give me a minute I'll figure it out if someone doesn't before me.

 

[anglum]

thanks a lot i really appreciate it.. i thought for sure the sin and cos of the angle would solve this for me... but it didnt and left me puzzled

 

[silvashadow]

http://img234.imageshack.us/img234/1231/26425932hu7.jpg
the box is applying 53.9sin34deg along the surface to the lower left
now we have that new triangle
cos 34=53.9sin34/F, solve for F
thats problem number 1 i think

 

[johnsonandrew]

Yeah definitely. Well I know the formula won't give you the answer becuase it finds what that dotted line equals, and we want the solid line labeled F. So I"m tryin to figure out how to do that...

 

[johnsonandrew]

Yeah there you go, but its 54.0(sin 34), if you get picky. No?

 

[johnsonandrew]

36.4 N for first answer?

 

[johnsonandrew]

I said (cos 34)= 30.19/F

 

[anglum]

when i solve for F for that i get 36.356N which is correct

now can i use that squared + X^2 = what squared to solve for number 2?

 

[silvashadow]

isnt it cos 34=30.1405/F?

 

[silvashadow]

yep, I am right and he solve problem 1

 

[anglum]

i got Number 1 now i am not sure how to solve for # 2 ?

 

[johnsonandrew]

I'm working on it..

 

[anglum]

would it be sin34 = 30.1405/F to get #2?

thanks guys u are a great help... this is seriously the hardest class ever to take strictly online

 

[silvashadow]

same concept:
Fn=53.9cos34, does return a wrong answer from the website?

 

[johnsonandrew]

Whattt? I got 1.45.. I did something wrong

 

[silvashadow]

the funny part is that I am learning this right now, like 2 days ago