Draw a diagram showing the forces acting on a plank and find x

[Richie Smash]

Homework Statement

Hello, it is me again, I recently posted about a question on moments and centre of gravity with two planks both 10 Kg in mass and 2m in lengths, uniform, situated on top of each other as shown in the picture.

The question asked for the maximum length of x that the system would be in equilibrium.
With much help, I now understand how to solve that part of the question, and I have done so.

The question now states, ''A 15 kg mass is placed at P. Draw a diagram showing the forces acting on the upper plank as it is about to topple. Use your diagram to find the value of x as it just begins to topple.''

Homework Equations

C.O.G= midpoint of uniform object
Moment= F *distance from line of action of force
Anticlockwise moments= clockwise moments in a system in equilibrium.

The Attempt at a Solution

So for the first part, it was seen that at the pivot point, the clockwise moments were 5x²/2

And for the anticlockwise moments they were (5x²-10lx+5l²)/2

With L being the length of the board, these moments were found by finding the centre of gravity of each side.
And the length of L was known to be 2, so ultimately it was solved that the maximum value of x for the system to stay in equilibrium was 1m

Now for this new part, I'm not sure how to represent this diagram, or how it's supposed to help me.

But I did try something, and what I did was, If 15kg is added t P, and they want the value of as it begins to topple , to me that sounds like while it just is about to exit equilibrium state,

And all I have to do is take these two equations
5x²/2=(5x²-10lx+5l²)/2

and add the 15 kg mass to the clockwise moments , because I know the mass per unit length is 5 kg.

That is how I got the 5x, because the mass on the right side would be 5*x

So All I'm thinking is to modify this and have 5x+15 because that would now be the total mass on the clockwise moment.

Now that I have the new mass, I can then multiply this mass (5x+15)* x/2 (the centre of gravity for the right side)

and my new moment will be (5x²+15x)/2

From there I just do this (5x²+15x)/2=(5x²-10lx+5l²)/2

According to the principle of moments, and since I know L to be 2m, I can solve for x which I found to be 0.6m.

Can anyone confirm if this is correct?

I will post the original picture and my own diagram that is required for the second part of this question.

 

[kuruman]

And the length of L was known to be 2, so ultimately it was solved that the maximum value of x for the system to stay in equilibrium was 1m ...

Which says that in the previous probem the centre of gravity of the top plank is right at the edge of the bottom plank. Where should the cg of the plank plus 15 kg mass be in this case for the plank to tip?

 

[Richie Smash]

I am thinking two things, It can be at the pivot point, or in the middle of the side of the plank that is hanging over.

 

[kuruman]

... or in the middle of the side of the plank that is hanging over.

Does this possibility make sense? That if you place extra weight at the tip of the plank it will tip at the same point as if there is no weight?

 

[Richie Smash]

NO that doesn't make sense, you're right, my intuition is telling me it would tip at the edge this time

I didnt want to put the value of gravity on my diagram, because I wasn't given it for this problem, I was able to solve part (i) with no gravity needed, i think what they are trying to stress here is centre of gravity, hence are continuously dealing with length, I want to know how to represent the Forces properly in my diagram.

 

[kuruman]

It always tips at the edge of the bottom plank. The question is at what distance x should the tip of the top plank be when the extra mass is added. Answer this: If without the mass x = L/2, when the mass is added, should x be greater or less than L/2? What does your intuition tell you?

 

[Richie Smash]

It would be less than, I took a 30cm ruler and I'm putting it at it's maximum equilibrium length 15cm, and then I shorten it to resemble X, and realize when I put my finger on the edge to act as the 15kg weight, the length of tipping would be shorter yes indeed

 

[kuruman]

Very good. Nothing is more convincing than the evidence before your eyes when you perform an experiment. The pivot point as you saw is the edge. Now gravity is an external force that acts at the cg of the top plank + 15 kg-mass. This means that, as far as gravity is concerned, you can replace the top plank + 15-kg mass with a point mass placed at the position of the cm. Draw two figures, one with the cg to the left of the edge and one with the cg to the right of the edge. What is the direction of the torque due to gravity in each of the pictures? What does this tell about the placement of the cg when the top plank + 15-kg mass when the system is just about to tip?

 

[Richie Smash]

Here is the right side with the direction of the torque

 

[Richie Smash]

Here is the left

 

[kuruman]

Excellent! So in view of the two drawings that show torques in opposite directions, where would you say the cg needs to be so that the torque is neither positive nor negative, i.e. zero?

 

[Richie Smash]

At the pivot point of the bottom plank?

Or if not I would say the mid point of those two individual Cg's

 

[kuruman]

Or if not I would say the mid point of those two individual Cg's

What do you think? Remember that you have replaced the top plank and the extra mass by a point 25-kg mass. There are no individual cg's. The bottom plank does not count because it's there to just provide the pivot at its edge.

 

[Richie Smash]

I'm finding this concept rather difficult to comprehend at times, but my intuition tells me that it would possibly be on the right side at the edge.

Or it tells me it might be on the far left side to counteract the force pushing it down... one of those two

How would the diagram they want to be drawn supposed to look? I don't have any given forces, so my best guess would be this point center of gravity and the upwards force of the plank, and the direction of the moment would be all the forces acting on the upper plank.

 

[kuruman]

I'm finding this concept rather difficult to comprehend at times, but my intuition tells me that it would possibly be on the right side at the edge.
Or it tells me it might be on the far left side to counteract the force pushing it down... one of those two

You need to make up your mind. However, we will abandon this line of reasoning for the time being and follow the way the problem is asking to go. Maybe when you reach the solution, you will see where I was going with this.

How would the diagram they want to be drawn supposed to look?

To draw the force diagram for the plank just as it is about to tip, you need to (a) consider all the forces acting on the plank and (b) draw these forces at the specific point on the plank where they act. So what do you think? How many forces act on the top plank and where do they act when the top plank is about to tip?

The drawing should show the top plank only with all the forces acting on it drawn in.

 

[Richie Smash]

I think this might be it

 

[kuruman]

Two of the three arrows you drew are incorrect.
1. The middle of the plank is the cg of the plank so the arrow in the middle of the plank is the weight of the plank only.
2. Where does the reaction force due to the bottom plank act when the top plank is just about to tip? Hint: What point of the bottom plank is in contact with the top plank when the top plank tips?

 

[Richie Smash]

I tweaked it a bit, it looks more correct to me now

 

[kuruman]

tweaked it a bit, it looks more correct to me now

You're still not there. Why is "the normal reaction force of lower plank" at the tip of the top plank? Remember that the top plank overhangs the lower plank, therefore there is no lower plank at that point to exert a reaction force.

 

[Richie Smash]

Oh now I finally get it,

 

[kuruman]

Super. Now just when the top plank is about to tip but hasn't tipped yet, what is the net torque acting on it?

 

[Richie Smash]

Sorry for the late reply, I think it would be this perhaps? (5x²+15x)/2=(5x²-10lx+5l²)/2

 

[kuruman]

Sorry for the late reply, I think it would be this perhaps? (5x²+15x)/2=(5x²-10lx+5l²)/2

At the moment of tipping the net torque is zero. I don't understand your expression. It looks like you are making this problem more complicated than it is. Below I show your figure redrawn to scale. You have 3 forces acting on the plank. What are the correct expressions for the torque generated by each force about the edge of the bottom plank at point O? Fill in the blanks with the appropriate expressions using symbols, not numbers. Use W_plank, N_reac., W_mass, x and L (= length of the plank).
1. Torque from weight of plank = __________ .
2. Torque from normal reaction force = __________ .
3. Torque from extra mass at the end of the plank = __________.

 

[Richie Smash]

Ok, got one

N_reac*(l-x)
W_mass*x
And for the plank...
W_plank*(l-x)

Because I believe that the reaction force and the plank force are counter clockwise moments

 

[kuruman]

You are not reading the figure correctly. Only the torque from W_mass is correct. What are the lever arms or distances from point O of N_reac. and W_plank. Remember that L or l is the total length of the plank.

 

[Richie Smash]

Ok the distance from point O for N_reac is 0
So it would be N_reac*0

The distance from W_plankthats... harder

 

[kuruman]

The distance from W_plankthats... harder

Well, say y is the distance that you are looking for. What is y + x in terms of L? Can you write an equation and solve for y?

 

[Richie Smash]

y= (l/2)-x

So:

W_plank*(l/2)-x

 

[kuruman]

Correct. Now add all the torques and require that they be equal to zero. Be sure to keep a consistent sign convention for clockwise and counterclockwise torques relative to point O.

On edit: More correctly your expression for the torque should be W_plank*[(L/2)-x], but I know what you mean.

 

[Richie Smash]

Ok I got,

W_m*x - W_p*((l/2)-x)+0=0

 

[kuruman]

W_m*x - W_p*((l/2)-x)+0=0

That is correct. What do you think you should do next?

 

[Richie Smash]

I am going to physics lessons now, will reply when I return

 

[Richie Smash]

Ok so I think I need to represent the Forces I have in terms of l and x, also, because I'm not sure i could just assume gravity is 10 and find the actual force...

So I would need to find the centre of gravity of the right side and then represent both the centre of gravity of the plank and the right side in terms of l and x, and solve for x since I know l = 2m

 

[kuruman]

You are making things more complicated than they are. You have this equation that you came up with, W_m*x - W_p*((l/2)-x)+0=0
Variable x represents the overhang that you are looking for. Can you solve the equation for what you are looking for? Note: It doesn't matter what value you use for g because it multiplies both terms on the left that add up to zero.

Note: After you find x, it would be instructive to find the position of the cg of top plank + extra mass relative to the right end of the top plank where the mass is. I have something to say about this that ties into my initial approach to guide you through this problem.

 

[Richie Smash]

Ok I have solved the equation as follows

used (z) to represent Wp, and (y) to represent Wm

yx-z((l/2)-x)=0
x-z((l/2)-x)=0
x-((l/2)-x)=0
x+x-(l/2)=0
(4x-l)/2=0
4x-l=0
4x=l
x=l/4

And since l =2

x=0.5m