Calculating mass of the horizontal leg of a L shaped lever

In summary, the problem involves finding the mass of the horizontal leg of a bell crank lever, given the masses and distances recorded during an experiment to bring it to equilibrium. The issue with the solution lies in the calculation of the clockwise and anti-clockwise moments, as well as the sum of forces. There seems to be a confusion in the units and distances used, as well as a mistake in switching the variables. Further clarification and assistance is needed to solve the problem.
  • #1
ridiculoid
19
0
Hey everyone, been stuck on this for over a day now, wondering if anyone is able to tell me if I am on the right track.

1. Homework Statement

"In your calculations for M1 (bell crank lever), note that the mass of the horizontal leg is contributing to the anti-clockwise moment. Given the masses and distances you have recorded for this experiment, calculate the mass of the horizontal leg."

http://imgur.com/a/ETYRT
Above is a diagram of what I am working with, I worked out the mass (0.3924N or 0.04kg) to bring it to equilibrium by physically attaching weights to the horizontal leg and now I am required to work out the mass of the horizontal leg itself. Also the circle of to the right is a pulley.

Homework Equations


clockwise moments = anticlockwise moments
F=mg
Sum of moments = 0
Sum of forces = 0

The Attempt at a Solution


I have attached my attempt in the form of a .pdf. I have tried to lay it all onto a horizontal plane like a beam support drawing because that is what I'm used to working with. The issue I am having is that my answer for the force coming down at the center of mass of the horizontal leg throws the balance of forces out, I end up with more downward force than upward force (which I have put at the pivot point).

Any help would be greatly appreciated.
 

Attachments

  • attempt.pdf
    63 KB · Views: 315
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  • #2
Just to clarify, we were required to bring this bell crank lever to equilibrium by attaching weights to it (it is pinned to a panel at the shoulder) and now with the information gathered, I need to find the mass of the horizontal leg of the bracket.

Ive also updated the imgur link to have my working on it as well.

I'm losing my mind over this!
 
  • #3
You know, I wanted to help out on this problem, but I just could not understand what was happening to be able to help. But I looked again and took some more time.

In summing the moments, for the 0.3924 N force, you multiply that force by a distance of 4. I don't understand what the units are, but it is a distance of 4. But for the 0.981 N force, you are also using a distance of 4. That is what you are using in the equation and that is the number of hash marks in your hand-drawn drawing. If I look at the other figure, for the 0.3924 N force, I can count that it is 4 blue circles away from point A. But for the 0.981 N force, I count that it is 3 blue circles away - not 4. I haven't worked the problem any further than that for now.

Edit: As I look further, you have:
2.3544 = F(2.5)
2.3544/2.5 = 0.9418
A = 0.9418.
Somehow, you switched the F to an A.

Then immediately after that, you wrote:
ΣM = 0 0.3924 + 0.981 - 0.9418 = 0.4316
F = 0.4316 N (with an arrow pointing in the down direction)
= 0.044 kg
You indicate that you are performing a sum of moments, but it is actually a sum of forces, ΣF = 0.
And it looks like it is a sum of vertical forces. But if that is the case (that it is a sum of vertical forces), then why are you adding in the 0.981 N? It looks like the pulley is independent of the L-shaped structure. Is that true, or not true? So it appears that the 0.981 N force is producing a horizontal force on the L-shaped structure - not a vertical force. Do you agree or disagree with that?

Edit 2: Welcome to Physics Forums.
 
Last edited:
  • #4
TomHart said:
You know, I wanted to help out on this problem, but I just could not understand what was happening to be able to help. But I looked again and took some more time.

In summing the moments, for the 0.3924 N force, you multiply that force by a distance of 4. I don't understand what the units are, but it is a distance of 4. But for the 0.981 N force, you are also using a distance of 4. That is what you are using in the equation and that is the number of hash marks in your hand-drawn drawing. If I look at the other figure, for the 0.3924 N force, I can count that it is 4 blue circles away from point A. But for the 0.981 N force, I count that it is 3 blue circles away - not 4. I haven't worked the problem any further than that for now.

Edit: As I look further, you have:
2.3544 = F(2.5)
2.3544/2.5 = 0.9418
A = 0.9418.
Somehow, you switched the F to an A.

Then immediately after that, you wrote:
ΣM = 0 0.3924 + 0.981 - 0.9418 = 0.4316
F = 0.4316 N (with an arrow pointing in the down direction)
= 0.044 kg
You indicate that you are performing a sum of moments, but it is actually a sum of forces, ΣF = 0.
And it looks like it is a sum of vertical forces. But if that is the case (that it is a sum of vertical forces), then why are you adding in the 0.981 N? It looks like the pulley is independent of the L-shaped structure. Is that true, or not true? So it appears that the 0.981 N force is producing a horizontal force on the L-shaped structure - not a vertical force. Do you agree or disagree with that?

Edit 2: Welcome to Physics Forums.

Yes I totally agree! I have messed this one up royally, those calculations were after a thought I had in the middle of the night and tried to hash out when I got to work but obviously my brain wasn't all there. I've also realized that because I need to find the weight of the horizontal leg, I can't just put in arbitrary numbers like we had been doing in class or I wouldn't get a 'real' answer.

Really sorry to have wasted your time, I've been going at it the last couple of hours and think I've made some progress.
 

Attachments

  • 20170403_193646.jpg
    20170403_193646.jpg
    19.4 KB · Views: 466
  • #5
ridiculoid said:
Yes I totally agree! I have messed this one up royally, those calculations were after a thought I had in the middle of the night and tried to hash out when I got to work but obviously my brain wasn't all there. I've also realized that because I need to find the weight of the horizontal leg, I can't just put in arbitrary numbers like we had been doing in class or I wouldn't get a 'real' answer.

Really sorry to have wasted your time, I've been going at it the last couple of hours and think I've made some progress.

By arbitrary I mean just counting the holes in the bracket as units, not using the measurements which I completely forgot I had taken, feel pretty stupid.
 
  • #6
ridiculoid said:
By arbitrary I mean just counting the holes in the bracket as units, not using the measurements which I completely forgot I had taken, feel pretty stupid.
Because the blue circles are so perfectly evenly spaced, you should have gotten the correct answer even if you had counted holes - provided you had used 3 instead of 4 when calculating the moment about A for the 0.981 force at the pulley. Because you are using the sum of moments equation to calculate a given force, it is only the ratio of the distances that matters - not the particular unit that you use. Your measured distance between holes turned out to be 20 (whatever the units are). And that means the force F should be at a perpendicular distance of (2.5 hole counts)x(20 units/hole count) = 50 units from point A.

You made a few mistakes, but it looks like your method is good.
 
  • #7
TomHart said:
Because the blue circles are so perfectly evenly spaced, you should have gotten the correct answer even if you had counted holes - provided you had used 3 instead of 4 when calculating the moment about A for the 0.981 force at the pulley. Because you are using the sum of moments equation to calculate a given force, it is only the ratio of the distances that matters - not the particular unit that you use. Your measured distance between holes turned out to be 20 (whatever the units are). And that means the force F should be at a perpendicular distance of (2.5 hole counts)x(20 units/hole count) = 50 units from point A.

You made a few mistakes, but it looks like your method is good.

Ah yes that does make sense, and I see where I've calculated that wrong. Thanks very much for your help.
 
  • #8
Questions:

Is the L bracket comprised of a single uniform thickness piece, or two pieces overlapped where they come together? (Perhaps two straight sections overlapped and welded?)

Are you to consider the "horizontal leg" to be the entire horizontal rectangular area, or just the portion that lies beyond the vertical leg edge (the non-overlapping area)? The difference has implications for the total mass of the "leg" and the mass distribution relative to the pivot point.
 
  • #9
gneill said:
Questions:

Is the L bracket comprised of a single uniform thickness piece, or two pieces overlapped where they come together? (Perhaps two straight sections overlapped and welded?)

Are you to consider the "horizontal leg" to be the entire horizontal rectangular area, or just the portion that lies beyond the vertical leg edge (the non-overlapping area)? The difference has implications for the total mass of the "leg" and the mass distribution relative to the pivot point.
Good point, they are 2 identical pieces overlapping and pinned together at the shoulder
 
  • #10
ridiculoid said:
Good point, they are 2 identical pieces overlapping and pinned together at the shoulder
So you'll need to make a decision about what the problem means by the label "lower leg", since where they overlap might be considered as either common to both legs, or not part of the legs, or part of one but not the other.

Can you give us your thoughts on what part of the bracket contributes to the net torque?
 

Related to Calculating mass of the horizontal leg of a L shaped lever

1. How do you calculate the mass of the horizontal leg of a L shaped lever?

To calculate the mass of the horizontal leg of a L shaped lever, you will need to know the length and density of the lever. Multiply the length by the density to find the mass of the lever.

2. What is the formula for calculating the mass of the horizontal leg of a L shaped lever?

The formula for calculating the mass of the horizontal leg of a L shaped lever is m = ρ * l, where m is the mass, ρ is the density, and l is the length of the lever.

3. What units should be used when calculating the mass of the horizontal leg of a L shaped lever?

The units used for calculating the mass of the horizontal leg of a L shaped lever should be consistent. For example, if the length is measured in meters, then the density should also be in kilograms per meter cubed (kg/m³) in order for the mass to be in kilograms (kg).

4. Can the mass of the horizontal leg of a L shaped lever vary?

Yes, the mass of the horizontal leg of a L shaped lever can vary depending on the length and density of the lever. A longer lever or a denser material will result in a higher mass.

5. How does the mass of the horizontal leg of a L shaped lever affect its balance?

The mass of the horizontal leg of a L shaped lever affects its balance by exerting a force on the fulcrum. A heavier mass on one side of the fulcrum will require a lighter mass on the other side to achieve balance.

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