# Bowling ball gutter with springs

1. Jan 7, 2009

### Mister V

1. The problem statement, all variables and given/known data

Edit:

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The give variables are:
Mass of bowling ball = 2 kg
Thèta (angle) = 30 degrees

The questions:
1. Draw the forces
2. Normal force of bowling ball on 1 plate
3. Force on springs

3. The attempt at a solution

The answer to the first question can be seen on the picture. I only drew the forces on one of the plate, since the forces on the other plate are symmetrical.
I hope its not too messy, but these are the forces:
Fs= force exerted by the spring on the plate, with Fs _|_ the right angle component on the plate.
Fn1= normal force exerted by the plate on bowling ball, and Fn'1= force exerted by bowling ball on the plate. (same for Fn2 and Fn'2).
Mg = weight of the bowling ball
Fo = force exerted by the "joint" on the plate.

Are the forces drawn correctly?

2nd question: normal force

Fn _|_ is the component of Fn in the vertical direction

As there are 3 forces in the vertical direction, I calculated Mg, the one pointing downward. Then I stated that the vertical components of the normal forces are the same, and 2 times on of those should be equal to Mg.

Mg = 2 kg * 9,81 m/s² = 19,62 N
cos 30 = (Fn _|_) / Fn
<=> Fn _|_ = Fn * cos 30

2 Fn _|_ = 19,62 N
so Fn_|_ = 9,81 N
so Fn = 11,33 N

Is this correct?

Third question

I took l to be the length of one plate.

You also now that:
(sin 30)/X = sin(60)/R where X = the distance from O to the Fn
So X = 0,58 R

Then choosing O as center:
- Fs _|_ * l + Fn * 0,58 R
<=> Fs _|_ = ( Fn * 0,58 R ) / l

Fs _|_ => cos 30 = Fs _|_ / Fs

So Fs = Fs _|_ / cos 30

Is this correct?

Is it possible to know the exact force on the spring? I would say it is impossible because you haven't got enough variables?

Last edited: Jan 7, 2009
2. Jan 8, 2009

### Mister V

No one? If I have to clarify something, or to add something, please say so. Thanks in advance

3. Jan 8, 2009

### Carid

Third Question
Your plate is operating as a lever about point O. If it is static then the two moment are equal.

[Moderation Note] Complete Solution Removed

Last edited by a moderator: Jan 8, 2009
4. Jan 8, 2009

### PhanthomJay

Are you sure that 'l' and 'R' are not given? I find it odd that a problem would assign a numerical value to M and Theta, but not give a value for R, and complete ignore l, the plate length. Otherwise, to get the spring force, you'd have to leave it as a function of l and R. I also find it difficult to identify forces acting on several different objects all on one diagram. It gets real confusing, especially with the different axes (x, y, and the one parallel to the plate) and Newton 3 to contend with.
Your attempt at a solution is very good, and you've done a lot of work in preparing the sketches, but I don't agree with your normal force calculation (the perpendicular force of the plate on the ball, or the ball on the plate). It appears to me that you have the weight force (mg/2) as a component of that normal force. Rather, it should be the other way around, that is, the Normal force is the component of the weight force. Also, note a couple of other things: the spring will handle resultant vertical forces only (I think you have correctly noted that) ; and your direction of the support reaction at O seems off...it will be the vector sum of the reactions from each plate.

5. Jan 8, 2009

### Mister V

I got Carid's answer by mail (the forum sends it automatically before it is edited). I know it is against the policy of the forum, but I would like to make some remark on it
You say Fn * R * sin(30°), and I guess thats force * position vector. I don't think this is right, because R * sin 30 is not the distance from O to the place where Fn 'starts'? Or am I overlooking something?

I also forgot to say that K is the spring constant, sorry about that.

They are not given. It is an exam question from last year, and I think it is this way ask the students to be calculate both with numbers and with symbols.

I'm sorry, I don't understand this. (English isn't my first language)

What I did was saying there are 3 forces important for the bowling ball. The weight, and the normal forces from the plates on the bowlingball.
I thought that, since the bowlingball is in an equilibrium, the resulting force should be zero.
So, in the vertical direction, the vertical components of the normal forces should be equal in magnitude to the weight.
So I stated that:
2 x (vertical component of the normal force) = Weight
And then you can calculate the normal force?

So, if I'm correct, Fo should be pointing upward?

6. Jan 8, 2009

### PhanthomJay

well, if they want to mix up the given variables with both letters and numbers, they at least should have the courtesy of denoting the plate length with a letter like 'l', rather than have you choose your own letter. But so be it.
You have calculated the normal force as $$mg/(2cos\theta) = 11.33N$$. What I am trying to say is that the normal force is $$(mgcos\theta)/2 = 8.5 N$$
Yes!

7. Jan 9, 2009

### Carid

Mister V wrote:
"You say Fn * R * sin(30°), and I guess thats force * position vector. I don't think this is right, because R * sin 30 is not the distance from O to the place where Fn 'starts'? Or am I overlooking something?"

Yes, my apologies, I made an error. The distance should have been R*sin(30°)/cos(30°)

8. Jan 9, 2009

### Mister V

Now I understand what you are trying to say, but I don't understand why it is that way?

If 8.5 N is the normal force, shouldn't both vertical components of the normal forces added together be equal in magnitude to Mg?

If 8.5 N = normal force
Then
Vertical component of normal force $$Fn*cos\theta$$
The value is then= 7,36
2 times the value of the vertical component = 14.72.

So the way I see I now, Is there is a force (the weight), in the vertical direction pointed downward with magnitude 19.62 N and 2 forces pointing upward with combined magnitude 14.72. So isn't there a net force pointing downward, which is in contrast to the situation of equilibrium?

Sorry to take much of your time with my probably stupid question...

9. Jan 9, 2009

### PhanthomJay

It is not a stupid question, as you may very well be correct; however, the more I look at this problem, the more I wish I hadn't. For example, assume theta, instead of being 30 degrees, was 89.9 degrees. This is a silly looking gutter, I know, but nonetheless, let's see what happens. The normal force becomes 9.81/cos89.9 = 5600N, and the Force in the spring becomes even greater than that, and the reaction at O becomes downward. It doesn't make any sense. Also, the spring must compress, but how? Perhaps you can shed more light on this than I.

Edit: I have come to the conclusion that at the given value of theta
- your solution for the normal force, and the spring force (as function of l and R) is 100 percent correct, and
- the spring must be allowed to translate horizontally at its base, to allow for the plate rotation as caused by the compressive downward displacement of the spring under the resultant vertical load on the spring. For small displacements , this doesn't change the solution significantly.

Nice job, I apologize for any confusion I may have caused.

Last edited: Jan 9, 2009
10. Jan 10, 2009

### Mister V

No problem, thanks for helping.

11. Jan 10, 2009

### PhanthomJay

Now I know why numerical values were given for M and Theta, because at large values of M and Theta, the spring forces become large, and the spring deformation becomes significant, causing large rotation of the plates, which changes the value of theta, and lowers the position of the ball, and the problem then becomes more complex and leads to a different solution. Anyway, excellent work!