Magnitude of Force within incline?

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Homework Help Overview

The discussion revolves around a physics problem involving a block on a frictionless incline, where participants are trying to determine the magnitude of a horizontal force and the normal force acting on the block. The incline is set at an angle of 34 degrees, and the mass of the block is given as 5.5 kg.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on the block, including the use of sine and cosine functions to resolve the forces along the incline and perpendicular to it. There are attempts to clarify the correct angles to use in calculations, with some confusion about the application of trigonometric functions. Questions arise regarding the accuracy of previous calculations and the significance of rounding in their answers.

Discussion Status

There is ongoing exploration of different approaches to solving the problem, with participants sharing their calculations and questioning the correctness of their results. Some participants suggest that the horizontal force does not equal the force parallel to the incline, indicating a need for further clarification. Multiple interpretations of the problem setup are being discussed, and guidance has been offered regarding the use of angles in calculations.

Contextual Notes

Participants note the potential impact of rounding on their answers and express uncertainty about the correctness of their calculations. There is mention of an online platform that provides immediate feedback on answers, which may influence their approach to solving the problem.

anglum
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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
 
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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.
 
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
 
silvashadow said:
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..
 
Is there additional information in the problem you didn't include? Because I believe that's correct...
 
those answers are incorrect... that's why i am so lost
 
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
 
those look correct to me too
 
i wish there was a way to draw out what they give u ... i tried to describe it the best i could
 
  • #10
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
 
  • #11
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
 
  • #12
and the horizontal comes from the left?
 
  • #13
i tried answers ranging from 30..05 to 30.2 and the same for 44.65 to 44.70

and they were wrong
 
  • #14
silvashadow said:
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...
 
  • #15
i am uploading the picture of the problem online.. i will have the link for u to see it in one minute
 
  • #16
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...
 
  • #18
anglum said:
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.
 
  • #19
check that link for the drawing of the problem
 
  • #20
Ok, give me a minute I'll figure it out if someone doesn't before me.
 
  • #21
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
 
  • #22
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
 
Last edited by a moderator:
  • #23
Yeah definitely. Well I know the formula won't give you the answer because 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...
 
  • #24
Yeah there you go, but its 54.0(sin 34), if you get picky. No?
 
Last edited:
  • #25
36.4 N for first answer?
 
  • #26
I said (cos 34)= 30.19/F
 
  • #27
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?
 
  • #28
isnt it cos 34=30.1405/F?
 
  • #29
yep, I am right and he solve problem 1
 
  • #30
i got Number 1 now i am not sure how to solve for # 2 ?
 

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