The discussion centers on calculating the upward force required to throw a 10,000 lb object 12 inches into the air using a platform that can only move vertically 6 inches. The participants emphasize the importance of understanding the mechanics involved, including the acceleration needed to achieve the desired height and the role of hydraulic rams in generating the necessary force. Key equations from Newton's laws of motion, such as \( u^2 = 2gh \) and \( u^2 = 2ad \), are highlighted as essential for solving the problem accurately.
PREREQUISITESEngineers, physics students, and anyone involved in mechanical design or dynamics who seeks to understand the forces involved in lifting heavy objects using limited travel mechanisms.
Is this problem for schoolwork? If not, what is the application? How do you "throw" a 10,000 pound object?theboom said:Say you have a platform, and on that platform you have an object the weighs 10,000lbs (ignore the weight of the platform itself). If the platform is only allowed to move vertically 6” before it must stop moving, how much upward force would be needed to throw the 10,000lb object 12” in the air? Also ignore air resistance.
It was a discussion that I saw recently but I changed the numbers. Obviously you can’t personally throw it but say you had some sort of mechanical device that was doing it, how much force would be needed? I changed the numbers because I wanted to see the effect of higher weight and limited travel distance of the force.berkeman said:Is this problem for schoolwork? If not, what is the application? How do you "throw" a 10,000 pound object?
theboom said:Say you have a platform, and on that platform you have an object the weighs 10,000lbs (ignore the weight of the platform itself). If the platform is only allowed to move vertically 6” before it must stop moving, how much upward force would be needed to throw the 10,000lb object 12” in the air? Also ignore air resistance.
I’ll put a method to it. Say the platform in sitting on a solid object at rest but underneath the platform is hydraulic rams. The stroke of the ram is only 6” so all of the upward force will have to be exerted on the object in the time it takes the ram to move 6” and within that 6” distance. How much would be needed to throw the object a further 12” into the air above the platforms final point (so 18” total above the original starting point, 12” above the platforms ending point).ZapperZ said:The question itself is incomplete. For example, what causes the platform to move? Is it supported by a spring? If so, what is the stiffness, or spring constant of the spring? If it is on concrete, why should it even move?
Secondly, how quickly do I have to lift the object? I can have the Hulk slowly lift it up 12" in the air, or I can put dynamites underneath the object and explode them, causing the object to jump up in the air very quickly. Each one of these scenario will produced different "impulse" and will cause different amount of reaction force being applied to the platform.
Zz.
theboom said:I’ll put a method to it. Say the platform in sitting on a solid object at rest but underneath the platform is hydraulic rams. The stroke of the ram is only 6” so all of the upward force will have to be exerted on the object in the time it takes the ram to move 6” and within that 6” distance. How much would be needed to throw the object a further 12” into the air above the platforms final point (so 18” total above the original starting point, 12” above the platforms ending point).
Don’t really understand your stiff question. When the platform is at its starting position, the hydraulic is not exerting any force on the platform so “zero stiff” I guess? Secondly, as far as speed, that’s part of my question. Obviously if you accelerate slowly the platform won’t be thrown at all. It will rise the 6” with the platform and stop when the platform does. I’m asking what force will be needed to generate the speed needed for a 10,000lb object to overcome the 9.8m/s/s of gravity enough to reach 12” of height before it starts coming back down.ZapperZ said:Again, how "stiff" is this hydraulics? And how fast do you want me to lift this? If I lift it very slowly, the platform will hardly move!
Zz.
Yes this is exactly right and what I’m trying to figure out.sophiecentaur said:I think this is doable. Work out the speed needed for the 6” throw. Then you need to accelerate the load over the bottom 6” at acceleration x to get that speed in 6”. The force needed will be (x+g)Times the mass.
There are other approaches but that way gives understandable (?) steps I think.
I assume you use the maximum lifting distance.
This is a fairly straightforward exercise using Newton's classical mechanics. Are you familiar with how the laws of motion work? d=st, v=at, etc?theboom said:Say you have a platform, and on that platform you have an object the weighs 10,000lbs (ignore the weight of the platform itself). If the platform is only allowed to move vertically 6” before it must stop moving, how much upward force would be needed to throw the 10,000lb object 12” in the air? Also ignore air resistance.
I’m somewhat familiar with them, although it’s been a bit since I practiced it. I’m assuming it’s higher than 9.8m/s? I can’t really remember.russ_watters said:This is a fairly straightforward exercise using Newton's classical mechanics. Are you familiar with how the laws of motion work? d=st, v=at, etc?
To start with, do you know how to calculate the required velocity? Here's a hint: It's the same on the way up as the way down.
You'll need to actually look up equations and calculate stuff here. Guessing or hoping someone gives you the answer isn't going to work. If you know the relationships between acceleration, speed and distance, post the equations, start manipulating them and see if you can work it out. We'll nudge you if necessary.theboom said:I’m somewhat familiar with them, although it’s been a bit since I practiced it. I’m assuming it’s higher than 9.8m/s? I can’t really remember.
theboom said:I’m somewhat familiar with them, although it’s been a bit since I practiced it. I’m assuming it’s higher than 9.8m/s? I can’t really remember.
No I don’t remember that one. Let me try that. I got 13.372mph..PeroK said:The first task is to calculate how fast the object must be traveling to go 12" (which is about 30cm in new money) into the air.
Does the equation ##u^2 = 2gh## look familiar?
theboom said:No I don’t remember that one. Let me try that. I got 13.372mph..