Maybe a silly question but on the above question using the conservation of momentum:
momentum before firing (0) = momentum after firing (55*35)+(M*2.5)
If I re-range the above it's M = -(55*35)/2.5 = -770kg. I can I reconcile that minus sign (basically get rid of it)?
Thanks
I mean what I found similar was the fact that you could take moments about any point on the ruler example e.g. at [0.2,0.8], [0.4.0.6] etc. and when you add the moments up they should always equal the couple...
The books solution says to take moments about x2 to find the force of x1 as 110N to 2 significant figures. Then by equilibrium: X1 + X2 = 250N therefore x2 = 250N - 110N = 140N.
I understood this to be (4/9)*250 = 110 2 s.f.
Similar to the reasoning of a couple i.e. (Couple = F*d) for...
Thanks Delta2. Sorry I got confused. I didn't think about the CoM/CoG where all the weight is acting. But now it's fine :-) The sum of the moments about the CoM should total 250N
Moment about X2 to calculate force at X1:
x1 * 9 = (250 * 2)
Therefore, x1 = 500/9 = 55.5N
The book however gives force at x1 as 110N. So I figured I have not understood a concept somewhere
In the attachment I am supposed to calculate the maximum frictional force of a block on a 35 degree angle incline (that is the point at which the force acting opposite frictional force is highest)
I make it out to be sin(35)*120N = 69N but the book says 675N and it gives it as...
Thanks! My point of confusion has always been around the difference between deceleration and negative acceleration. Deceleration as you said is slowing down and negative acceleration just depends on what we take as positive or negative direction I guess
Hi All,
Please see attached photo of the question.
It is asking for net force and acceleration. Taking the forces acting upwards and downwards on the
parachutist as vectors: for A the net force would be 800-300=500; B would be 0 and C; would that be 800-1500=-700? And if so would that imply...