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

[newbphysic]

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.

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 ?

 

[ehild]

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.

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 F_A and F_B. Solve for F_B in terms of the angle q. At what q is F_B minimum?

 

[newbphysic]

Write the equations for the x and y components of the forces F_A and F_B. Solve for F_B in terms of the angle q. At what q is F_B minimum?

F_Ax = F_Acos(20)
F_Ay = F_Asin(20)F_Bx = F_Bcos(q)
F_By = F_Bsin(q)

can you help me with q ?
i know q must make
F_By=-F_Ay
F_Bx + F_Ax = 950
But i don't know how to find it.

 

[ehild]

Plug in the expressions with the angles for the force components.

 

[newbphysic]

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]

Now eliminate F_A.

 

[newbphysic]

Now eliminate F_A.

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]

 

[ehild]

Isolate F_B.

 

[newbphysic]

[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]

Isolate F_B.

 

[ehild]

Fb is function of q and you need to find the minimum of that function. Have you studied Calculus?

 

[newbphysic]

Fb is function of q and you need to find the minimum of that function. Have you studied Calculus?

only calculus 1

 

[ehild]

What is the derivative of a function at a minimum or maximum?

 

[newbphysic]

What is the derivative of a function at a minimum or maximum?

zero

 

[ehild]

Take the derivative of Fb with respect to q.

 

[newbphysic]

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]

That expression would be equal to zero. You will be able to solve that equation for q.

 

[newbphysic]

That expression would be equal to zero. You will be able to solve that equation for q.

The answer :

hmm,it looks like mission impossible . Do you know another method ehild ?

 

[ehild]

It is a fraction equal to zero. What do you know about the numerator, if the fraction is zero?

 

[newbphysic]

numerator must be zero which means q must be 0 for sine and 90 for cosine

It is a fraction equal to zero. What do you know about the numerator, if the fraction is zero?

 

[ehild]

q is the same angle both for sine and cosine. What equation do you have for q?

 

[newbphysic]

my guess q must be 135(-225) or 315(-45) because sin and cos have different sign there

q is the same angle both for sine and cosine. What equation do you have for q?

 

[ehild]

Do not ques, solve. What is the equation?

 

[newbphysic]

Do not ques, solve. What is the equation?

950 sin(q) + 2626.47 cos(q) = 0

 

[ehild]

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).

 

[newbphysic]

All right, thanks a lot ehild.
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]

I don't think it is true. It must be an other problem. What did you get for q? Are F_A and F_B perpendicular ?

 

[newbphysic]

I don't think it is true. It must be an other problem. What did you get for q? Are F_A and F_B perpendicular ?

if i use that method the result is the same. q = 70 degrees clockwise from x.

 

[ehild]

That method is not correct for the original problem.

 

[newbphysic]

That method is not correct for the original problem.

so it's just a coincidence that it has the same result ?

 

[ehild]

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?

 

[newbphysic]

The result are not same, only close together. What did you get with my method?

70.11 clockwise

 

[ehild]

It should be 71.12 clockwise.

 

[newbphysic]

950 sin(q) + 2626.47 cos(q) = 0
950 tan (q) = -2626.47
tan q = - 2626.47 / 950
q =

It should be 71.12 clockwise.

 

[ehild]

There was some mistakes when we derived F_B, and I did not notice.
Next time do the derivation symbolically, and plug in numerical data at the end. In that case, you would have got 70°.

It is true that the two angles add up to 90°.
If the angle of F_A is a and the angle of F_B is q, and the magnitude of the sum of the forces is F
F_Acos(a) + F_Bcos(q) =F, (we measure q clockwise),
F_A sin(a) = F_B sin(q) --> F_A=F_B sin(q)/sin(a).
[tex]F_B=\frac{F}{\frac{\sin(q)}{\tan(a)}+\cos(q)}[/tex]

If you want F_B minumum, the denominator should be maximum. Take the derivative of ##\frac{\sin(q)}{\tan(a)}+\cos(q)## and make it equal to zero : ##\frac{\cos(q)}{\tan(a)}-\sin(q)=0##, that is ##\frac{1}{\tan(a)}-\tan(q)=0##
What does it mean for the angles a and q?

 

[newbphysic]

There was some mistakes when we derived F_B, and I did not notice.
Next time do the derivation symbolically, and plug in numerical data at the end. In that case, you would have got 70°.

It is true that the two angles add up to 90°.
If the angle of F_A is a and the angle of F_B is q, and the magnitude of the sum of the forces is F
F_Acos(a) + F_Bcos(q) =F, (we measure q clockwise),
F_A sin(a) = F_B sin(q) --> F_A=F_B sin(q)/sin(a).
[tex]F_B=\frac{F}{\frac{\sin(q)}{\tan(a)}+\cos(q)}[/tex]

If you want F_B minumum, the denominator should be maximum. Take the derivative of ##\frac{\sin(q)}{\tan(a)}+\cos(q)## and make it equal to zero : ##\frac{\cos(q)}{\tan(a)}-\sin(q)=0##, that is ##\frac{1}{\tan(a)}-\tan(q)=0##
What does it mean for the angles a and q?

For F_B perpendicular F_A,is there an explanation for this ? why it works ? and why not perpendicular to F_R ?

Sorry if i ask a lot of questions :)