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

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Homework Help Overview

The problem involves optimizing the forces in a towing scenario, where a truck is to be towed using two ropes. The objective is to determine the magnitudes of the forces acting on each rope and the angle of one of the forces to minimize its magnitude while maintaining a resultant force of 950 N directed along the positive x-axis. The angle of one force is fixed at 20° from the x-axis.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to write equations for the x and y components of the forces and express one force in terms of the angle of the other. There are attempts to isolate variables and derive conditions for minimizing one of the forces. Questions arise about the calculus involved in finding minima and the implications of angles in the context of the problem.

Discussion Status

The discussion is active, with participants exploring different mathematical approaches to derive the necessary conditions for minimizing the force. Some participants have suggested using calculus to find the derivative of the force function, while others are questioning the assumptions about angles and relationships between the forces. There is no explicit consensus on the final approach or solution yet.

Contextual Notes

Participants note that there are only two known values, which complicates the use of certain laws of trigonometry. There is also mention of potential mistakes in earlier derivations and the need for careful symbolic manipulation before substituting numerical values.

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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.
 
Plug in the expressions with the angles for the force components.
 
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
 
Now eliminate FA.
 
ehild said:
Now eliminate FA.
Fb cos(q) + 0.94 Fa = 950
Fb sin(q) + 0.34 Fa = 0
--------------------------------------------- -

\frac{Fbcos(q)}{0.94} - \frac{Fbsin(q)}{0.34}= \frac{950}{0.94}
 
Isolate FB.
 
Fb ( \frac{cos(q)}{0.94} - \frac{sin(q)}{0.34})= \frac{950}{0.94}

Fb = \frac{950}{0.94} / (\frac{cos(q)}{0.94} - \frac{sin(q)}{0.34})
ehild said:
Isolate FB.
 
  • #10
Fb is function of q and you need to find the minimum of that function. Have you studied Calculus?
 
  • #11
ehild said:
Fb is function of q and you need to find the minimum of that function. Have you studied Calculus?
only calculus 1
 
  • #12
What is the derivative of a function at a minimum or maximum?
 
  • #13
ehild said:
What is the derivative of a function at a minimum or maximum?
zero
 
  • #14
Take the derivative of Fb with respect to q.
 
  • #15
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?
 
  • #16
That expression would be equal to zero. You will be able to solve that equation for q.
 
  • #17
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 ?
 
  • #18
It is a fraction equal to zero. What do you know about the numerator, if the fraction is zero?
 
  • #19
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?
 
  • #20
q is the same angle both for sine and cosine. What equation do you have for q?
 
  • #21
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?
 
  • #22
Do not ques, solve. What is the equation?
 
  • #23
ehild said:
Do not ques, solve. What is the equation?
950 sin(q) + 2626.47 cos(q) = 0
 
  • #24
newbphysic said:
950 sin(q) + 2626.47 cos(q) = 0
OK. divide the equation by cos(q). Replace sin(q)/cos(q) by tan(q). Solve for tan(q).
 
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  • #25
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 ?
 
  • #26
I don't think it is true. It must be an other problem. What did you get for q? Are FA and FB perpendicular ?
 
  • #27
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:
  • #28
That method is not correct for the original problem.
 
  • #29
ehild said:
That method is not correct for the original problem.
so it's just a coincidence that it has the same result ?
 
  • #30
newbphysic said:
so it's just a coincidence that it has the same result ?
The result are not same, only close together. What did you get with my method?
 

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