# Force needed to move an object from one point to another.

by sofcal
Tags: force, object, point
 P: 1,345 Hello sofcal! If you're talking about moving this cylinder horizontally along the ground then the force required to move the cylinder must over come the force of friction: $$Moving force > \mu_{kinetic} mg$$ Where mu is the kinetic friction coefficient (a constant determined by the two surfaces that meet: in your case the ground and the bottom of the cylinder). m is the mass of the cylinder and g is the acceleration due to gravity ~9.81 on Earth. If you're talking about moving the cylinder straight up into the air it must over come the force of gravity: $$force upward > mg$$ I'm not particularly sure if this is what you were looking for but perhaps it helps... Oh I should also remark that a force accelerates an object so you will have to take that into consideration as well. In case you don't know how the two are related: $$F = \frac{d(mv)}{dt} = m \frac{dv}{dt} = ma$$
 P: 1,345 Force needed to move an object from one point to another. Yes you are correct but there are a few things I should warn you about: 1.) s = vt - (at^2)/2 assumes there is a constant acceleration and by extension a constant force 2.) Be careful about your sign! Acceleration is a vector so make sure in s = vt - (at^2)/2 a has the right sign (positive a for this particular equation would be slowing down, while negative a would be speeding up) 3.) A force changes the acceleration so if you want to make something stop faster apply a negative force for a longer time, or a larger force for a short period of time. Another helpful equation might be: $$Impulse = F \Delta t = m \Delta v$$ So let's run through this: If you have an object just floating in the air (no friction) and you want to move it horizontally you apply a (constant) force which will (constantly) accelerate an object. When that force is released the object will continue to move with a certain constant velocity (because there is no longer acceleration). To stop the object you must apply a (negative) force so the object will de-accelerate until the objects velocity is 0 (v = 0 implies no more movement obviously) Make sense?
 PF Gold P: 864 Hi sofcal. You seem to be having several problems. First, you need to be able to calculate the position and velocity with a given force. The equations you are using give an exact solution for a constant acceleration. The moment the force changes, they stop working. They are not the most convenient form to use in a computer. There are many algorithms people use in computers to do accurate simulations of physical systems (look up the verlet alogorithm if you are interested) but that is overkill here, I think. I think you just want to make the motions seem natural to the user (If I am not mistaken) and in this case, Euler's method is very straightfoward and will give good enough results. Here is how you use it: At some time, t, the object has position x(t), velocity v(t), and the force on it is F(t). You want these variables at the next time step, $$t + \Delta t$$. Since acceleration is the rate at which velocity changes, a good approximation for small time steps is: $$v(t+\Delta t) = v(t) + a(t)*\Delta t$$ $$F = ma$$, so $$a(t) = \frac{F(t)}{m}$$ so the velocity at the next time is $$v(t+\Delta t) = v(t) + \frac{F(t)*\Delta t}{m}$$ For the position at the next time you can use $$x(t+\Delta t) = x(t) + v(t)*\Delta t$$ or, alternatively $$x(t+\Delta t) = x(t) + \frac{v(t)+ v(t+\Delta t)}{2}*\Delta t$$ These two should be almost equal for small $$\Delta t$$ You just apply this iteratively, using the results of one step to put into the next.
 PF Gold P: 864 The next problem you will run into is what to use for the force. If you simply apply a constant force on the object in the direction of the mouse, or if you use a force proportional to the distance between the two, you will have "ringing" This means the object will overshoot the mouse, keep going until it slows down completely and comes back, then overshoots again, and so on forever. You need to add "damping". This means that part of the force just opposes the velocity and tends to bring the object to a rest. I would suggest a force that looks like this: $$F(t) = -k*(x(t) - x_{mouse}(t)) - c*v(t)$$ The undefined symbols are: $$x_{mouse}(t) =$$ position of the mouse at time t k = "Spring" constant, force of attraction between object and mouse c = "Viscous damping coefficient", force that slows the object down You can experiment with different values of 'k' and 'c'. Different values will give different behavior. See http://en.wikipedia.org/wiki/Damping