Problem statements
Solution video - H17.A
Solution video - H17.B
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HOMEWORK 17 - Fall 24
22 thoughts on “HOMEWORK 17 - Fall 24”
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Does anyone have any idea how the weight influences the answer for the maximum ratio of b/2h? I may have taken the wrong approach, but based on how I solved for the ratio it seems like weight doesn't influence the result. Logically though I would think that the weight would affect it somehow?
I also ended up getting an answer where weight doesn't influence the result. The way I thought about it is that if the beam weighs more, then there is an proportional increase in the normal force as well. As a result, the force of friction also increases proportionally, meaning that it can hold the increased weight.
I agree with Jackson. But I'll elaborate a bit more in my reasoning. I think that based on my math, because you're trying to create an equilibrium system, you're trying to equal to friction quantities, so the weight, whatever it might be will always cancel out between them because it will always influence each side the same relative to one another, causing it to not matter.
Yes once you substitute into moment equation you will see that weight can be factored out and therefore does not influence the answer. Logically you would think it would but due to friction being present on both surfaces it does not.
I believe the weight does not affect the final answer because it will directly cause a proportional increase in normal force and also the friction force giving you the same answer. You could also find this by doing algebra and canceling out variables in an equation where everything is put together.
I believe that algebraically, the weight would not influence the answer. However, after the weight crosses a certain limit, the static friction does not increase along with the weight, rather it turns into dynamic friction. I wonder if that affects the answer after the weight has crossed the critical point.
When solving for the largest ratio of b/2h that the ladder can sustain static equilibrium, taking the moments around different points of where static friction occurs the weight is the same value in both equations which points to Weight not having an effect on the ratio of b/2h.
Just thinking about the problem intuitively too, the weight of the object shouldn't have an effect on the final answer.
For example, if you had a ladder resting vertically, you could theoretically make the ladder infinitely heavy and the ladder wouldn't tip over unless some sideways force was applied to create a moment.
This problem really centers around keeping the moment about the center of mass in equilibrium. If the friction forces keep the moment as 0, then the ladder won't move.
For solving this problem, do you want to take the moment around multiple points and then solve for the ratio of b/2h? This is the method I have thought about but have been a bit confused of how to solve this.
I think you can solve it with Fx, Fy, and a moment about one of the ends of the rod, plugging things in and cancelling until you get the ratio.
When solving this problem is it better to have the coordinate plane set up where the x-axis is parallel to the ladder and the y-axis is perpendicular to the ladder, or should the x and y axis be set up where the x-axis is parallel with the ground and the y-axis would be perpendicular with the ground?
You may have an easier time setting up the coordinate plane so that it is in the direction where x and y are parallel to the wall/ground and not the ladder. This is because most of the forces (friction, normal, and weight) of the beam fit better in this coordinate plane.
Does the force of gravity affect the system or is that the same thing as the normal force of the weight?
the force of gravity just indicates the direction of the weight of the ladder (W) as vertically downwards. This should act at G, the center of gravity of the ladder. Hope that helps!
By taking the moments, will my weight and normal force cancel out leaving me with a way to calculate the ratio, or is there a different way to go about solving
The weight will cancel out, and the normal forces will be substituted using the equations that you get by equating all horizontal and vertical forces respectively to zero.
Do we have to solve for weight using the normal forces of y and x?
You do not need to numerically solve for the weight, but it should be included in your force balance equations and will be a variable you carry around until it gets canceled out.
What variables should the final ratio of b/2h be in terms of? I'm getting multiple possible answers and don't know what we should be looking for.
b/2h should only be in terms of numbers. All of the other variables should factor and then cancel out.
I originally was confused as well, but in the end I found that after proper setup all the variables canceled and you are left with only a number/decimal for your final b/2h.
Wouldn't the weight matter at some point, theoretically? Because if you think about it, if theladder's weight is taken as a limit to infinity, then it would make sense that it would eventually slide and break the coefficient of friction.