Optimizing Forces in Towing: Minimizing FB for a Given Resultant Force

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newbphysic
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Homework Statement


The truck is to be towed using two ropes. If the
resultant force is to be 950 N, directed along the positive x
axis, determine the magnitudes of forces FA and FB acting
on each rope and the angle q of FB so that the magnitude of
FB is a minimum. FA acts at 20° from the x-axis as shown.

UgXVI2g.png


Homework Equations

The Attempt at a Solution


I have no idea where to start because there is only 2 known values so i can't use sine law or cosine law. What's the first step ?
 
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newbphysic said:

Homework Statement


The truck is to be towed using two ropes. If the
resultant force is to be 950 N, directed along the positive x
axis, determine the magnitudes of forces FA and FB acting
on each rope and the angle q of FB so that the magnitude of
FB is a minimum. FA acts at 20° from the x-axis as shown.

UgXVI2g.png


Homework Equations

The Attempt at a Solution


I have no idea where to start because there is only 2 known values so i can't use sine law or cosine law. What's the first step ?

Write the equations for the x and y components of the forces FA and FB. Solve for FB in terms of the angle q. At what q is FB minimum?
 
ehild said:
Write the equations for the x and y components of the forces FA and FB. Solve for FB in terms of the angle q. At what q is FB minimum?

FAx = FAcos(20)
FAy = FAsin(20)FBx = FBcos(q)
FBy = FBsin(q)

can you help me with q ?
i know q must make
FBy=-FAy
FBx + FAx = 950
But i don't know how to find it.
 
ehild said:
Plug in the expressions with the angles for the force components.
FBx + FAx = 950
FBcos(q) + FAcos(20) = 950
FB cos(q) + 0.94FA = 950

FBy=-FAy
FB sin(q) = - 0.34 FA
 
ehild said:
Now eliminate FA.
Fb cos(q) + 0.94 Fa = 950
Fb sin(q) + 0.34 Fa = 0
--------------------------------------------- -

[tex]\frac{Fbcos(q)}{0.94} - \frac{Fbsin(q)}{0.34}= \frac{950}{0.94}[/tex]
 
[tex]Fb ( \frac{cos(q)}{0.94} - \frac{sin(q)}{0.34})= \frac{950}{0.94}[/tex]

[tex]Fb = \frac{950}{0.94} / (\frac{cos(q)}{0.94} - \frac{sin(q)}{0.34})[/tex]
ehild said:
Isolate FB.
 
ehild said:
Fb is function of q and you need to find the minimum of that function. Have you studied Calculus?
only calculus 1
 
ehild said:
What is the derivative of a function at a minimum or maximum?
zero
 
ehild said:
Take the derivative of Fb with respect to q.
derivative of cos and sin will result another sin and cos
what should i do with it?
 
ehild said:
That expression would be equal to zero. You will be able to solve that equation for q.

The answer : :nb)
CI5PCPg.png

hmm,it looks like mission impossible . Do you know another method ehild ?
 
numerator must be zero which means q must be 0 for sine and 90 for cosine
ehild said:
It is a fraction equal to zero. What do you know about the numerator, if the fraction is zero?
 
my guess q must be 135(-225) or 315(-45) because sin and cos have different sign there
ehild said:
q is the same angle both for sine and cosine. What equation do you have for q?
 
ehild said:
Do not ques, solve. What is the equation?
950 sin(q) + 2626.47 cos(q) = 0
 
All right, thanks a lot ehild. :smile::smile::smile:
One more question though, there is other method that says in order for Fb to be minimum it must perpendicular to Fa ? Can you explain why it works ?
 
ehild said:
I don't think it is true. It must be an other problem. What did you get for q? Are FA and FB perpendicular ?
if i use that method the result is the same. q = 70 degrees clockwise from x.
 
Last edited:
ehild said:
That method is not correct for the original problem.
so it's just a coincidence that it has the same result ?
 

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