Determine force angles so that acceleration is least

In summary, for (c), if you change the angles of F2 and F3 so that their X-components cancel out f1, then the resultant force is the smallest possible, which is zero.
  • #1
MrMoose
23
0

Homework Statement



Figure below is an overhead view of a 12kg particle that is to be pulled by three ropes. One force (F1, with magnitude 50N) is indicated. Orient the other two forces F2 and F3 so that the magnitude of the resulting acceleration of the particle is least, and find that magnitude if

(a) F2 = 30N, F3 = 20N
(b) F2 = 30N, F3 = 10N
(c) F2 = F3 = 30N.

Figure: 0----->F1

Homework Equations



Ftx = F1 + F2x + F3x = ma

a = (F1 + F2x + F3x) / m

F2x = F2 * cos(θ)

F3x = F3 * cos(θ)

The Attempt at a Solution



Initially, I said to myself, "well the least the acceleration can be is zero" so I assumed a = 0, Ftx = 0 and it follows that F1 = -F2x - F3x. Therefore:

F1 = -F2 * cos(θ) - F3 * cos (θ)

Solving for θ;

θ = Arcos[F1 / (-F2x - F3x)]

This gives you the angle of F2 and F3 and you've already made the assumption that a = 0

This worked out for parts (a) and (c) of the problem, but it wasn't until I dug into part (b) that I saw a few flaws in my logic:

1) The problem doesn't say that acceleration is zero, it says to find the angle for the forces so that the acceleration is least. The only problem is, how do I set up my formula so that mathematically, the difference between the sum (F2x + F3x) and F1 is as small as possible?

2) My final formula for θ assumes that both F2 and F3 have the same direction (in other words, θ2 = θ3), which may not be true. But if I introduce θ2 and θ3 into my equations, they become very difficult to solve.

Thanks in advance, MrMoose
 
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  • #2
Remember that your acceleration is in TWO directions, the x and y directions. Thus,the magnitude of the acceleration is the "hypotenuse" of the "acceleration triangle".

Let us take set up the problem:

We have:
F_1+F_2cos(a)+F_3cos(b)=ma_x
F_2sin(a)+F_3sin(b)=ma_y

We now square both equations, sum them together, in order to find the equation for the squared magnitude of the acceleration:
We gain (remember sin^2t+cos^2t=1 for all t!):
F_1^2+F_2^2+F_3^2+2F_2F_3(cos(a)cos(b)+sin(a)sin(b))+2F1(F_2cos(a)+F_3cos(b))=m^2(a_x^2+a_y^2)
-------------------------------
Now, this problem IS solvable, using partial derivatives, but I do not think that was the intention.
Rather, I believe either you are the book has left out the critical information that a_y=0!Am I correct about that?
 
  • #3
Okay, I made this too difficult.
In a) and b), the answers must necessarily be to place the two other vectors in opposite direction to F_1

But, in c), no acceleration in y-direction yields the condition:
30sin(v_2)+30sin(v_3)=0, meaning v_2=-v_3

Insert these values in the x-equation
 
  • #4
Yup, now I see what you did. Conceptually, you ask yourself, "how can I make the total force, Ft, as small as possible?" For (a) and (b), the sum of the forces f2 + f3 < f1. They will always produce the smallest Ft on the -x axis. However, in (c), f2 + f3 < f1, and you know that the x components of f2 and f3 will decrease as their angles increase. Now you have an opportunity to change the angles of f2 and f3 so their X components cancel out f1, yielding the smallest Ft, which is zero. In that case, I can use my assumptions from the original attempt (especially because f2 = f3 and you know θ2 = θ3). Thanks!
 
  • #5
Actually, v_2=-v_3, but it doesn't matter for the even cosine functions.
If I did this right, you ought to get the equation cos(v_2)=-5/6, for zero acceleration in x-direction as well.
 

1. How do you determine force angles to minimize acceleration?

To determine force angles that result in the least amount of acceleration, you must use the principles of vector addition. This involves breaking down the forces acting on an object into their individual components and finding the resultant force. The angle of this resultant force will determine the direction of acceleration, and adjusting the angles of the individual forces can minimize this acceleration.

2. What factors affect the force angles for minimizing acceleration?

The main factors that affect the force angles for minimizing acceleration are the magnitude and direction of the individual forces acting on the object. The angles at which these forces are applied also play a crucial role in determining the resulting acceleration. Additionally, the mass and shape of the object, as well as external factors like friction, can also impact the force angles.

3. Is there a specific formula for determining force angles to minimize acceleration?

There is no specific formula for determining force angles to minimize acceleration, as it depends on the specific scenario and the forces involved. However, the principles of vector addition and trigonometry can be used to find the resultant force and its angle, which can then be used to determine the desired force angles.

4. Can force angles ever completely eliminate acceleration?

No, force angles cannot completely eliminate acceleration. According to Newton's second law of motion, an object will always experience acceleration when there is a net force acting on it. However, by manipulating the force angles, we can minimize the acceleration and potentially achieve a state of equilibrium where the net force is zero.

5. How can understanding force angles and acceleration be useful in real-world applications?

Understanding force angles and acceleration is crucial in various fields, including physics, engineering, and sports. In physics and engineering, it is used to design structures and machines that can withstand external forces and minimize acceleration. In sports, athletes can use knowledge of force angles to optimize their movements and improve their performance.

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