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
--------------------------------------------- -

\frac{Fbcos(q)}{0.94} - \frac{Fbsin(q)}{0.34}= \frac{950}{0.94}

 

[ehild]

Isolate F_B.

 

[newbphysic]

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})

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).
F_B=\frac{F}{\frac{\sin(q)}{\tan(a)}+\cos(q)}

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).
F_B=\frac{F}{\frac{\sin(q)}{\tan(a)}+\cos(q)}

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 :)

 

[haruspex]

For FB perpendicular FA,is there an explanation for this ? why it works ?

It's easy to see geometrically.
Draw the 950N as a line OP with a length representing its magnitude, and a line for the direction of F_A from the same origin.
The magnitude of F_A will be indicated by a point on its line, Q say.
F_B will be represented by a vector from Q to P. What position of Q minimises the length PQ?

 

[ehild]

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 :)

It IS the explanation. $\vec F_R$ is the sum of the vectors $\vec F_A$ and $\vec F_B$. We expressed their components with the magnitudes F_A, F_B and angles, a, q. Solved the system of equations for F_B. It must be minimum, so its derivative with respect to q has to be zero. From that, we got the equation $\frac{1}{\tan(a)}-\tan(q)=0$, that is, the tangent of the angle q is equal the reciprocal of the tangent of the other angle. Now, remember how the tangent was defined in a right triangle,

$tan(\alpha)=a/b$ and $tan(\beta)=b/a$, the tangents of the two angles are reciprocal to each others in a right triangle. But the sum of the two angles is 90°as the sum of all angles in a triangle makes 180°, one angle is 90° , so $\alpha + \beta = 90°$. Any time when the tangents of two acute angles are reciprocal, their sum is 90°.
Here, the two forces $\vec F_A$ and $\vec F_A$ enclose the angle a+q, and the tangents of the angles are reciprocal to each other, so a+q=90°, the forces are perpendicular to each other.

 

[newbphysic]

It's easy to see geometrically.
Draw the 950N as a line OP with a length representing its magnitude, and a line for the direction of F_A from the same origin.
The magnitude of F_A will be indicated by a point on its line, Q say.
F_B will be represented by a vector from Q to P. What position of Q minimises the length PQ?

when it's perpendicular to F_R ? the red line is the right answer according to calculation

$tan(\alpha)=a/b$ and $tan(\beta)=b/a$, the tangents of the two angles are reciprocal to each others in a right triangle. But the sum of the two angles is 90°as the sum of all angles in a triangle makes 180°, one angle is 90° , so $\alpha + \beta = 90°$. Any time when the tangents of two acute angles are reciprocal, their sum is 90°.
Here, the two forces $\vec F_A$ and $\vec F_A$ enclose the angle a+q, and the tangents of the angles are reciprocal to each other, so a+q=90°, the forces are perpendicular to each other.

OK but what i still don't understand is this, how can you know it must be perpendicular to F_A not to F_R. Here is another example that looks exactly like before but in this example Fb perpendicular Fr. Why it's different ?

I tried to use the your method and i find theta = 60 degrees

 

[ehild]

OK but what i still don't understand is this, how can you know it must be perpendicular to F_A not to F_R. Here is another example that looks exactly like before but in this example Fb perpendicular Fr. Why it's different ?

I tried to use the your method and i find theta = 60 degrees

This is an other problem. Here Fa is given, FR has to be determined, so as its direction is the positive x axis. In the previous problem, FR was given and Fa was unknown.
From the equation Fa sin(30)=Fbsin(θ) ---> 2*1/2 = Fbsin(θ), you get that Fb=1/sin(θ), which is minimum if sin(θ)=1, that is, θ=90°.

 

[newbphysic]

This is an other problem. Here Fa is given, FR has to be determined, so as its direction is the positive x axis. In the previous problem, FR was given and Fa was unknown.
From the equation Fa sin(30)=Fbsin(θ) ---> 2*1/2 = Fbsin(θ), you get that Fb=1/sin(θ), which is minimum if sin(θ)=1, that is, θ=90°.

I'm really confused

Why for the first problem i can't solve it like this one (without substituting F_A)
F_A sin(20) -F_B sin(q) = 0
F_B =F_Asin(20)/sin(q)

I know that we don't know F_A but we know in order for F_B to be minimum sin(q) must equal 1 (the biggest value) but the result will not be 70

 

[haruspex]

when it's perpendicular to F_R ? the red line is the right answer according to calculation

Neither line minimises the distance PQ, keeping P fixed and varying Q. (Or is P supposed to be where the red line meets the baseline?)

 

[newbphysic]

Neither line minimises the distance PQ, keeping P fixed and varying Q. (Or is P supposed to be where the red line meets the baseline?)

P is fixed but Q vary.

Ooh i see so is this mean in order to find minimum value the unknown must perpendicular to each other ?

 

[haruspex]

P is fixed but Q vary.

Ooh i see so is this mean in order to find minimum value the unknown must perpendicular to each other ?

If one has a fixed direction and you are trying to minimise the other, yes.
(I assume Q in the diagram is where the blue line meets the sloping black line.)

 

[ehild]

I'm really confused

Why for the first problem i can't solve it like this one (without substituting F_A)
F_A sin(20) -F_B sin(q) = 0
F_B =F_Asin(20)/sin(q)

I know that we don't know F_A but we know in order for F_B to be minimum sin(q) must equal 1 (the biggest value) but the result will not be 70

In the first problem, FR was given, you knew the point P and you had to draw a line to the line of FA, which length was the shortest. It is the normal from P to the line FA. That line segment corresponded to FB, and the red line segment corresponded to FA.
In the second problem, FA was given, you knew the point Q. You have to draw the shortest line from it to the line of FR. It is perpendicular to FR.
You know, that the shortest distance from a point to a straight line is the perpendicular to the given line.

If you drew a line perpendicular from P, its length would be longer than the real FB.

 

[newbphysic]

If one has a fixed direction and you are trying to minimise the other, yes.
(I assume Q in the diagram is where the blue line meets the sloping black line.)

In the first problem, FR was given, you knew the point P and you had to draw a line to the line of FA, which length was the shortest. It is the normal from P to the line FA. That line segment corresponded to FB, and the red line segment corresponded to FA.
In the second problem, FA was given, you knew the point Q. You have to draw the shortest line from it to the line of FR. It is perpendicular to FR.
You know, that the shortest distance from a point to a straight line is the perpendicular to the given line.
View attachment 97245

If you drew a line perpendicular from P, its length would be longer than the real FB.
View attachment 97246

Thank you for your help