Recent content by Torater

Thank you so much for all your help, and putting up with my dumb errors throughout hahah

T= 8(9.81sin30 - 3.02) T= 15.1N? Before I did this: T=8(9.81sin30 - 3.02)/sin30

so then my tension should work out to 30.2... haha let's hope I did this right

(and your arithmetic in your "long approach" was wrong, so it's difficult to check the rest) Where did I go wrong so I can correct this... Like are my sum of force equations right? Did I substitute them into other equations wrong? Or is the fact that I didn't use the Y forces at all...

oh well thanks! haha the last 48hr's of physics frustration can do that to a person ok attempt 2 at shortcut: pseudo mass: mr^2 / r^2 = 5 M = 5+5 = 10 a= (8(9.81) sin30/10 a= 3.92 Ok let's focus on the long way as its what the question asks I don't think I did that...

ya i'll have to do it the other way because it specifies using Newtons laws but I can confirm I didn't mess up my algebra by this handy method!

tiny tim i did my algebra wrong! 3.74??

r= .25m m1=5kg I=mr2 M=5+cos30(5) mt=7.5 + 8 a= (8 (9.81)sin30/15.5 a= 2.53

Ok take a look at my problem that has incline in the heading Ill go try it right now with this trick!

gneil does this work for problems with inclines as well?? I have posted a question with an incline, if this trick works for that too that is very handy!

∑fy=m2a = N-m2gcos30 ∑fx=-m2a = Tsin30-m2gsin30 ∑τ= Iα = Tr T= m2(gsin30-a)/sin30 I=mr² m1a=m2(gsin30-a)/sin30 sin30 (5 a)= (8 (9.81sin30)-a) sin30 (5a) + 8a = 39.24 a= 6.04m/s² T= (8)(9.81sin30-6.04)/sin30 T= -18.2N

An 8kg mass is initally at rest on a 30 degree frictionless incline, it is attached to a hoop shaped 5kg pulley of radius 25.0cm by a massless cord. Find the tension in the cord and the acceleration of the mass using Newton's Laws.

Ok let's try this again: ∑τ=Iα = Tr ∑Fy=-m2a = T-m2g I= 1/2mr²Therefore 1/2m1r²(a/r)= m2(g-a) (r) 1/2m1 a = m2(9.81-a) 1/2 (8) (a) = 5(9.81-a) a= 5.45m/s² T= (5(9.81-5.45)) T= 21.8N ?I hope your not in my class CHOD! lol

whoops I put that into the old equation so my acceleration is correction which I then can solve for T as: 5(9.81-2.25) T= 37.8N

2/5m1r²(a/r) = (m2(g-a) - m1a) (.5) Okay I took my sum of Torque equation and substituted equations 1 and 2 so that I only have one unknown variable :a I did this like so: ∑τ= Iα = Fr sin 90 Iα= (2/5m1r²)(α) which alpha is = to (a/r) and set this equal to F r so I have: (2/5m1r²)(a/r) = Fr...