4/18/2023 0 Comments Weight physics calculator![]() Your final statement is incorrect (regardless of typos and other errors) if you accelerate 100 kg down at 8 m/s^2, you will still have to resist gravity (push up) with 181 kg*m/s^2 (Newtons). Assume up is positive for force, displacement, velocity and acceleration. Second, once you pick a direction as positive, you must to stick with it. Recalculate all acceleration rates based on these numbers (deceleration rates will be identical in magnitude, opposite in sign). For the 0.5 second raise (or lower), assume the accel time to be 0.25 sec, and the accel distance to be 0.5 m. To minimize your accelerations, assume that accel time and decel time are equal (accel and decel distances are also equal). There must be a deceleration portion to each move. It takes F5/F3=5.35 times as much force to do so! That's 5.35 times more force to go slow than fast!įirst, you can't assume that you can accelerate 100% of the way up or down. ***Contrary to your belief, it takes MORE force to lower the weight 1m in 4s than it does to lower it in 0.5s. Now compare it to the force required to lower 100kg, 1m, in 4sį3=968.5 kg m/s^2 is required to lower the weight 1m in 4s However, now let's look at what happens on the way down.Ĭalculate the force required to lower 100kg, 1m, in 0.5s:į3=181 kg m/s^2 is required to lower the weight 1m in 0.5s You're right about this but nobody is disagreeing with you here It takes F1/F2=1.73 times as much force to do so. ***As predicted, it takes more force to raise the weight 1m in 0.5s than it does to raise it in 2s. Now compare it to the force required to raise 100kg, 1m, in 2sį2=1031 kg m/s^2 required to raise the weight 1m in 2s Now let's solve for the forces required to accelerate the weightĬalculate the force required to raise 100kg, 1m, in 0.5s:į1=1781 kg m/s^2 required to raise the weight 1m in 0.5s Known: Mass(m) =100kg (220lbs) Acceleration(a)=? Distance(d)=1mįirst let's solve for the acceleration required to move the weight 1m in the time frames of the sets.Ĭalculate a, to travel 1m in 0.5s (Raising and Lowering of set 1):Ĭalculate a, to travel 1m in 2s (Raising set 2):Ĭalculate a, to travel 1m in 4s (Lowering set 2): For simplicity's sake we'll assume the weight accelerates 100% of the way up and down for both sets. Let's assume we're comparing 2 sets using the same weight, but different rep speeds. Thx in advance for your time if anyone is able to work it out, here are the calculations a friend worked out, hope they are right not sure. It’s nothing to do with homework by the way, I have posted similar questions before, it’s just for a debate we are having. ![]() However, as the force need is going to be higher in the transition from the eccentric to the concentric, in each repetition of the lifting and lowering in weightlifting, I was hoping to work this out, as it’s a bit over my head. Someone worked out for me the lifting and lowering of a 100kg for 1m each way at. 5 of a second, in 1m, and at the last instant has the accelerate this weight up 1m at. However the lifter first is lowering this weight at. ![]() I was trying to work out the force needed to lift a 100kg weight. ![]()
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