Physics Help Forum Force to move object

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 Jul 25th 2009, 02:21 AM #1 Junior Member   Join Date: Jul 2009 Posts: 1 Force to move object I am attempting to build a tilting deck - a square frame supported by a centre post which can tilt horizontally/vertically - but I cannot work out the Force required to move the deck vertically up/down. In a simple 2D drawing, the centre post would be 300mm from ground to top, and the deck would extend 600mm right and 600mm left (a 'T' shape). If a person of 100KG stood on the deck 300mm from the centre, how much force would be required to move the deck vertically upward 150mm (at the outside edge) in 2 secs - and again, down 150mm in 2secs? How does the position of the applied force change the amount of force required? eg. if the force is applied 600mm from the centre vs 300mm from the centre? I would prefer the formula to make the calculations myself rather than answers so I tune the system to work correctly. PS: This is not school work (I think I remember school.. was quite a while ago now) but I failed miserably at physics during high school (because I was useless at it ) The reason I am asking is that I need to purchase Linear Actuators to perform the movements but they are rated in Newtons and I cannot work out which formulas to use to convert the Newton's to Work. Thanks in advance. Tony Some workings I have done: Force * ForceArm = Resistance Force * Resistance Force Arm x * 0.6 = 0.3 * 100Kg x = (0.3 * 100) / 0.6 x = 50Kg or 490.5 Newtons From this I assume that it requires 490.5N to hold the deck level when the weight is applied. To move vertically 150mm (0.15m) in 2 seconds.... A = 2*D / T*T A = 2 * 0.15m / 2sec * 2sec A = 0.3 / 4 A = 0.075m/s/s F = MA F = 100KG * 0.075 F = 7.5N Does this mean the force to lift the deck upward 150mm from level in 2 seconds is 490.5+7.5=498N and to lower it 150mm from level in 2 seconds is 490.5-7.5=483N?? Last edited by sentientnz; Jul 25th 2009 at 02:57 AM.
 Aug 1st 2009, 02:30 AM #2 Junior Member   Join Date: Jul 2009 Posts: 15 close, but the acceleration is not .075, that is the speed. .15m/2s= .075m/s. v=at^2 or .075=a(2)^2= a4 so a=.075/4 or .0188m/s^2... by my calculations, anyway.

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