Recent content by sandman203

Homework Statement Hi all this is just a quickie, but in one of my texts a question asks to find the maximum constant deceleration given an initial speed, time to stop and the distance it took. Now I am not sure whether to use v = u + at or s = 1/2at^2 + ut im thinking more v = u + at...

wow ok finally haha. guess we will see when it gets marked! thanks once again for the help dude realllllllllllly appreciate it!

so NA = 7×g - Fsin30 so FfA = 0.4 × (7×g - Fsin30) Now Fx = m×ax = Fcos30 - FfA From before ax = 2.803 & FfA = 0.6 × (7×g - Fsin30) so Fx = 7×2.803 = Fcos30 - ((0.4 × 7 × g) - (0.4 × Fsin30) Fx = 19.621 = Fcos30 - ((0.4 × 7 × g) - (0.4 × Fsin30) Fx = 19.621 + 27.44 = Fcos30 +...

Ah ok, yes the question does not state that block A is at rest. I am going to go with the 0.4 co-efficient because u guys are way smarter than me. But can acceleration not imply that a body is being accelerated from rest? Anyways thanks a **** tonne for the help TSny, must of been a pain dealing...

hmmmm 0.6? aren't we trying to overcome the static friction? I am not sure how i'd incorporate Kinetic friction because isn't the question asking the force needed to make the block slide from rest?

Ok so NA = 7×g - Fsin30 so FfA = 0.6 × (7×g - Fsin30) Now Fx = m×ax = Fcos30 - FfA From before ax = 2.803 & FfA = 0.6 × (7×g - Fsin30) so Fx = 7×2.803 = Fcos30 - ((0.6 × 7 × g) - (0.6 × Fsin30) Fx = 19.621 = Fcos30 - ((0.6 × 7 × g) - (0.6 × Fsin30) Fx = 19.621 +...

Ok so the x component of Wy = 0. So fxb = 2*ax = -17.481sin15 + 10.4886 cos15 so ax = 2.803m/s^2 & then F = 52.126N

Ok so Nb = 17.481 Fxb = 2*-ax = 17.481cos15 -2gsin15 + 10.4886cos15 so ax = -4.0422 Now subbing into fxab F = 59.88N

Ok so using the above method i get Nb = 17.481, subbing into fxb i get ax = 2.529m/s^2 then subbing into Fxab i get 50.48N

why is it not Nb, the normal is perpendicular to the incline surface? 2×axsin15 is the vertical component of F acting on B? or does only the horizontal portion of F work on B? how does friction come into action at all on the y-axis? I've changed the co-ord system when dealing with B. should i...

ok so I've changed ƩFyb = 0 = NB + 2×axsin15 - mb×g which decreases the normal.. i understand that now.. i didnt understand that before because i was thinking that only the x portion of the Force applied was acting on B. Now by realising the Y component of F also acts on B i can see how it...

ahh no it has none, otherwise it would lift off the block or go into block A. right?

ok so for block b ƩFy = 2×-gcos15 = -Fsin15 + NAB +2×axsin15??

Im having trouble seeing how the acceleration to the left cause less normal force can someone please explain this to me, i really feel like I am going insane over this. please help!

Ok guys, last attempt at answering this Q before i /wrists SO here are the 4 equations after applying F Block AB: ƩFyAB = F sin 30 + NAB - mAB×g ∴ NAB= mAB×g - F sin30 ƩFxAB = - Fcos30 +FfA = - mAB×ax Block B: ƩFxB = 2×-axcos15 = 2×g×sin15 - Ffb ƩFyb = 0 = NB - 2×axsin15...