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Eng physics - help needed

  • Thread starter 9danny
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  • #1
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eng physics - help needed!!

I have three simple problems that - even though I understand - I can't get the answer the book is saying...
1) Two forces with same magnitude F. What is the angle between the vectors if their sum has a magnitude of (sqrt)2F.

** ok - I'm doing R = (sqrt)2F therefore R = (sqrt)(Fx^2 + Fy^2) but Fx = Fy
Then R = sqrt(2F^2)
solving for F --> F= R/2 --> F= (sqrt2)/2

I'm doing (angle) = cos^-1 {[(sqrt2)/2]/[sqrt2]} but I get 60!! Book is telling me it should be 90deg... **


2) Two horses pull horizontally on ropes attached to a tree stump. The two forces F1 and F2 are such that the net force R has a magnitude equal to that of F1 and makes an angle of 90 with F1. Let F1=1300N. Find the magnitude of F2 and its direction.

** The book is answering 1840N and 135 deg. I broke R into Rx and Ry, same for F1...
Rx=(1300N)cos90 = 0 Ry=(1300N)sin90=1300 etc
Then using both components for both forces F1 and R - using Rx = F1x + F2x and solving for F2x (and F2y later) and again Pitagoreas' theorem I got ::
F2 = sqrt[ F2x^2 + F2y^2 ]
which - strange - gives me zero!! What did I do wrong? **
 

Answers and Replies

  • #2
Doc Al
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9danny said:
1) Two forces with same magnitude F. What is the angle between the vectors if their sum has a magnitude of (sqrt)2F.

** ok - I'm doing R = (sqrt)2F therefore R = (sqrt)(Fx^2 + Fy^2) but Fx = Fy
Then R = sqrt(2F^2)
solving for F --> F= R/2 --> F= (sqrt2)/2

I'm doing (angle) = cos^-1 {[(sqrt2)/2]/[sqrt2]} but I get 60!! Book is telling me it should be 90deg... **
Not sure what you are doing here. Instead, let the first vector be F in the x-direction. Let the second vector be F at some angle [itex]\theta[/itex]. Now find the resultant by adding the components of the two vectors. Apply what you know about the magnitude of the resultant to solve for [itex]\theta[/itex].


2) Two horses pull horizontally on ropes attached to a tree stump. The two forces F1 and F2 are such that the net force R has a magnitude equal to that of F1 and makes an angle of 90 with F1. Let F1=1300N. Find the magnitude of F2 and its direction.

** The book is answering 1840N and 135 deg. I broke R into Rx and Ry, same for F1...
Rx=(1300N)cos90 = 0 Ry=(1300N)sin90=1300 etc
Then using both components for both forces F1 and R - using Rx = F1x + F2x and solving for F2x (and F2y later) and again Pitagoreas' theorem I got ::
F2 = sqrt[ F2x^2 + F2y^2 ]
which - strange - gives me zero!! What did I do wrong?**
Nothing wrong with this method, but realize that F1y = 0.
 
  • #3
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The first one is definitely 90 just draw a triangle with two legs of equal side F and with a hypotenuse of Fsqrt2 you should just be able to look at that one and evaluate it. Remember put the vectors head to tail and their sum (magnitude) is the vector that goes from the tail of the first to the head of the second.
 
  • #4
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How can I show my work?
Fx/R = cos(angle) then solve for angle gives me 45deg
Fy/R = sin(angle) gives me 45deg -- then what? add them?
 
  • #5
Doc Al
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9danny said:
How can I show my work?
Fx/R = cos(angle) then solve for angle gives me 45deg
Fy/R = sin(angle) gives me 45deg -- then what? add them?
What problem are you referring to?
 
  • #6
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Doc Al said:
What problem are you referring to?

The first one.
The resultant = sqrt2 F
HELP
 
  • #7
Doc Al
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9danny said:
How can I show my work?
Fx/R = cos(angle) then solve for angle gives me 45deg
Fy/R = sin(angle) gives me 45deg -- then what? add them?
If you are referring to problem #1, R is the resultant of both vectors.

F2x = F2 cos (angle)
F2y = F2 sin (angle)

These are both true, but not especially helpful.

Instead, find the components of each vector and add them:

[tex]R_x = F + F\cos \theta[/tex]
[tex]R_y = F\sin \theta[/tex]

Now apply
[tex]R^2 = R_x^2 + R_y^2[/tex]
 
  • #8
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Doc Al said:
If you are referring to problem #1, R is the resultant of both vectors.

F2x = F2 cos (angle)
F2y = F2 sin (angle)

These are both true, but not especially helpful.

Instead, find the components of each vector and add them:

[tex]R_x = F + F\cos \theta[/tex]
[tex]R_y = F\sin \theta[/tex]

Now apply
[tex]R^2 = R_x^2 + R_y^2[/tex]

Is R_x^2 = F_x^2??
Where am I solving for the angle?
What is F?

From
[tex]R^2 = R_x^2 + R_y^2[/tex][/QUOTE]
I find that R_x = R_y... therefore R^2 = 2R'^2
Solving --> R' = (sqrt)1
Am I right?
 
  • #9
Doc Al
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I'm still not sure what you are doing. I gave you what Rx & Ry are. Plug them into the equation I gave for [itex]R^2[/itex]. (Actually do the squaring!) Since you know that [itex]R^2 = 2 F^2[/itex] (that's given), plug that into the equation. Then you can solve for the angle.
 

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