Recent content by liamke70

the instantaneous centre of zero velocity(sorry should of mentioned that) the red line is connecting the dots of the instantaneous centre of zero velocity at five positions through the movement of the pliers the top link (top handle) is fixed in my example dose this clear it up? Thank you :)

trying to plot the movement of the instantaneous centre of lock jaw pliers first image is an example of the 4 bar link and the second is a sketch which I've drawn of what i think it would be. is this the only instantaneous centre?

sorry i meant joules im still not seeing how it would work, how would the equation be written?

okay so say the work done to the spring is 6 N then using the net force kx - ma i could make the equation (6-kx)/m=-a is this right?

i don't see how i can use Newtons second law f=ma to solve for acceleration in this case a= f/m cant be applied as I am dealing with energy and work

i have used energy methods(potential energy to work done to the spring) and got to the max compression so now what do i do to get the acceleration at maximum compression? rearrange the net force (kx -mg) to get acceleration?

or is it rather kx/m=a x= deforamtion k= spring rate

okay so if energy causing the spring to compress say 6 J(Ep) and its maximum deflection is 0.02m then how would the acceleration be calculated? maybe x being deflection length 6= max 6/mx=a

f=ma , a=f/m forces and balances f=0 therefore a = 0 is that right?

well there's no change in velocity over time so i guess there's none i think I'm just paranoid that my lecturer is trying to trick me haha

Thank you :) while i have you here, could you please answer one more thing? if a spring is at its point of maximum deflection, does it the body the spring is attached to have any acceleration? the body is landing on the springs

okay, so your saying the 100N of thrust doesn't alter the mass but just the net force, and then net force x h (energy) that becomes kinetic energy. so then can i then just put the energy(net force x h) and mass(50kg) into E= 1/2mv^2 to get the velocity? Thank you very much for your replies

how would i include upwards thrust?

Just wondering if apparent weight affects kinetic energy or is it only mass that affects it regardless of apparent weight? for example, a helicopter is falling but is providing a small thrust. say the helicopters mass is 50kg and it provides 100N of thrust upwards therefore, using w=mg, the...