# Weight needed to launch object to a certain height

by njr140
Tags: height, launch, object, weight
 P: 2 Hi, I'm a first time poster here and I have a physics question I can't find the answer to anywhere. I don't know much about physics so this might be a basic question to answer. I have a mass (x) that is attached to a cable that is attached to a counterweight that is at 10 feet that will be dropped to ground level via a pulley at the top. The cable will release x once it reaches the end of the launch track and the object will go vertical with the maxium amount of kinetic energy. The launch would occur horizontally and the object would then go vertical. X needs to reach a certain height (z) (10 vertical feet). I don't know how fast the mass would be moving; it would depend on the speed of the counterweight dropping freely with gravity. As of now, I don't need friction or air resistance factored in, just the weight. The question I have is, if you have a variable mass (x), how much counterweight (y) will you need to launch the object 10 feet (z)? I have created a GIF to give a visual since it's kind of hard to explain what I mean. http://s27.postimg.org/ml4madh83/output_xd_Ooc_F.gif
 Homework Sci Advisor HW Helper Thanks P: 12,794 Welcome to PF; Use conservation of energy to get a minimum weight needed. If you want to launch mass m directly upwards to a height h, that takes energy mgh where g is the acceleration of gravity. This energy is being supplied by mass x falling distance y ... it's the same equation: xgy. In your case, y=h, so x=m. This is with no friction or other losses.
 P: 933 How much do we know about the length of the cord that connects mass X to mass Y? As drawn, it seems that the cord length is greater than 20 feet. If that is correct then what is the speed of mass X at the moment that mass Y touches down? In particular, is that speed adequate to propel mass X 10 feet in the air?
 P: 2 Weight needed to launch object to a certain height Yes, the cord would be greater than 20 feet. The speed is unknown. Thats what I'm trying to figure out. I need to know the weight of y would be to provide whatever speed is needed to launch x to 10 feet.
 Sci Advisor PF Gold P: 9,437 It is complicated by not knowing the mass density of the tether.
Homework
HW Helper
Thanks
P: 12,794
 I need to know the weight of y would be to provide whatever speed is needed to launch x to 10 feet.
... helps to be consistent though.

if you want to throw mass x, vertically, to a height of 10', using the energy from dropping a mass y by 10', then you need at least y=x.

How much more mass y you need depends on the tether mass, friction, and other losses.
Weigh the tether (rope? chain? string?) and divide that weight by the length.
You can also try different masses for y and measure how high x gets - plot a graph.
 Sci Advisor Thanks PF Gold P: 12,135 If you are trying to design an alternative to a trebuchet, you need to bear in mind that it could be pretty inefficient - i.e. the mechanical energy supplied by the falling mass will probably need to be severals times greater than the kinetic energy that actually gets delivered to the projectile. To be 100% efficient, the falling mass should end up stationary, having delivered all its energy (that's before it hits the bottom of its travel). Perhaps the falling mass should have a curved track, too. I notice that your X travels a lot further than your Y (?). In fact, looking at your diagram, there is no Velocity Multiplication so X will never end up higher than 10+10 ft. You must have a lever, gears or some pulleys to make X travel faster than Y so it can go higher than Y falls. Look at Trebuchet designs - very effective before gunpowder came along.
P: 933
 Quote by njr140 Yes, the cord would be greater than 20 feet. The speed is unknown. Thats what I'm trying to figure out. I need to know the weight of y would be to provide whatever speed is needed to launch x to 10 feet.
Given this clarification, it is impossible. x will never even reach the top of the launch track.

The velocity ($v_y$) of y when it touches down will be something less than $\sqrt{2gh}$. The velocity ($v_x$) of x at that instant will be the same. The height that x can reach (measured from the ground) is then given by $\frac{{v_x}^2}{2g}$.

That's $\frac{something less than\sqrt{2gh}^2}{2g}$ which will be less than h.
 Sci Advisor Thanks PF Gold P: 12,135 @njr I think you have done a thought experiment and you have not actually included the basics of mechanics in it. That's very easy to do if you do not have the 'Physicist Habit'. The basic principle is that you can never get more energy out than you put in; mostly you get much less. For your projectile to reach a certain height from where it started, it will need a minimum of mgh energy. If this energy comes from giving it an initial upward velocity then that velocity must be √2gh. A falling mass, from the same height could transfer some of its kinetic energy to you projectile but it cannot easily give it enough, without, as I wrote before, some lever or gearing. Your question is too open and, so far, your model will not really work. Where do you want to start, to improve on it?

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