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 May 3rd 2017, 03:25 AM #31 Junior Member   Join Date: May 2014 Posts: 18 Hi Chip, This now leads me to counter weights. Would you know a formula in which I can change the following values and play around with different weights and time: Let's say I want to lift 8500 pounds 100 feet into the air. Let's say the permanent counter weight is 8400 pounds and that I add 2000 pounds to the counter weight side lifting the 8500 pound weight 100 feet. I would like to know how long it would take for the 8500 pound weight to rise 100 feet. Thank you again for all of your help, I am most grateful.
 May 3rd 2017, 08:00 AM #32 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,281 For two masses on a pulley system, M_1 and M_2, using $\displaystyle \sum \vec F = m \vec a$ you have: $\displaystyle (M_2 - M_1)g = (M_1 + M_2)a$ This gives you a value for the acceleration of the masses. The time it takes to move a certain distance 'd' given initial velocity = 0 comes from: $\displaystyle d = \frac 1 2 a t^2$ Solve for t.
 May 18th 2017, 04:49 AM #33 Junior Member   Join Date: May 2014 Posts: 18 Hi Chip, Not sure how to ask this, but here it goes. Suppose you have a pulley with a rope. On one side you have a four pound weight and on the other side a three pound weight. Obviously the four pound weight would pull the 3 pound weight up. But what if the 4 pound weight was in the shape of a ball which had a hole drilled into with a rod going through it so you could attach the rope to it and instead of dropping the ball staight down you rolled the ball down say a 15 degree slope. Do you lose any of the energy, in other words would the 3 pound weight rise more slowly rolling the 4 pound weight down a slope as opposed to dropping it straight down? (Pretty sure you would). And if so, by how much? Is there a formula to figure out different weights and slopes? Cheers, Kenji
 May 18th 2017, 06:39 AM #34 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,281 If the system is frictionless there is no loss of energy, BUT the masses will definitely accelerate slower. The reason is two-fold: First, the component of the force of gravity that is pulling on the 4-pound ball in the the direction of the ramp is smaller than if the 4-pound weight is allowed to fall straight down. In fact, at some angle that force will precisely equal the force of gravity acting on the 3-pound weight, and the system will no accelerate at all. And at lower ramp angles the system will run backward - the 3-pound weight will cause the 4-pound ball to roll uphill. The "magic angle" for this is arcsin(m/M) = 48.6 degrees. This comes from $\displaystyle \sum F = ma$: $\displaystyle Mg \sin \theta - mg = (M+m) a$ Here M is the mass of the large ball, m is the mass of the smaller weight, and 'a' is the acceleration of the system. Positive values of 'a' means the ball is rolling down the ramp and the 3 pound weight is rising. If the left hand side = 0 then 'a' = 0 and the system is in equilibrium, and if the left hand side is negative then 'a' is negative. Second, in order for the 4-pound ball to roll (as opposed to slide down the ramp), some of the potential energy that is converted to kinetic energy is used to spin the ball (i.e it goes into rotational KE) and some is used to get the masses to moving linearaly: $\displaystyle \Delta KE + \Delta PE = 0$ $\displaystyle Mg h \sin \theta - mgh = \frac 1 2 Mv^2 + \frac 1 2 m v^2 + \frac 1 2 I \omega^2$ Given that $\displaystyle \omega = v/R$: $\displaystyle (M \sin \theta)gh = \frac 1 2 v^2(M+m+\frac I {R^2})$ Here 'h' is the vertical displacement upward of the 3-pound weight. From this you can calculate velocity as a function of displacement of the weights. This can be simplified given that for a rolling ball $\displaystyle \omega = v/R$, and for a solid sphere $\displaystyle I = \frac 2 5 MR^2$: $\displaystyle (M\sin \theta + m)gh = \frac 1 2 v^2(\frac {7M} 5 + m)$
 May 3rd 2018, 12:10 PM #35 Junior Member   Join Date: May 2014 Posts: 18 Gravity Generator Hi Chip, It has been a while. Hope all is well. I wanted to see if I am understanding buoyancy correctly. Premise: I have an empty cylinder 4 feet in diameter and 10 feet in length floating on the surface of the water. I wish to force the cylinder to a depth of 100 feet using a weight. For this example it can be an anchor. I estimate the weight of the anchor needs to be at least 7844 lbs. to pull the cylinder down to a depth of 100 feet. Is this correct? Also I am estimating that you would need 126 cu/ft of air in order to float that 7844 lbs. anchor to the surface. Is this correct? If these are not correct, do you have a formula that will be able to answer these two premises. Thank you. Cheers.
 May 3rd 2018, 02:12 PM #36 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,281 Hello Tonk, Your calculations are correct, though you are assuming the weight of the cylinder and the air inside has no influence. If the dimensions you are given are the external dimensions of the cylinder, then the weight of the anchor should be reduced by the weight of the cylinder plus any air inside. As for floating the cylinder - you stated that it is “empty” to begin with - do you mean it’s filled with air? Or do you mean it’s a vaccuum, and you’re wondering if you need to insert air into the vaccuum to make the cylinder float back up to the surface?
 May 3rd 2018, 02:29 PM #37 Junior Member   Join Date: May 2014 Posts: 18 Gravity Generator Hi Chip, I meant if you were to open a valve and fill the cylinder with water at 100 feet, then close the valve and then using the 126+ cu/ft of air to float it to the surface with the 7844 pound anchor still attached. Cheers, Kenji

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