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post the figure or a diagram.
i have T-102Nsin(40)-19.5=m1a =-6.09
but i know this is wrong, i think i'm supposed to use the total weight of the system after watching some youtube videos
block on ramp: T -102N*cos(40) - uk*102N*sin(40) = (102N/g)*a suspended block: 32N-T =(32N/g)*a solve for T T=32N -(32N/g)*a plug into first equation and solve for a.
i think i keep confusing my cos and sin
hold up. I think I may have confused sin and cos. lol.
yeah I switched them, sorry: block on ramp: T -102N*sin(40) - uk*102N*cos(40) = (102N/g)*a
this isn't different than what i am getting then my numbers are 24N-65.6N-19.5N divided by 10.4
how'd you find T?
i thought we drew that angle like this|dw:1348867283133:dw| i got a = 2.45 solving the frictionless problem then solved fro t in x2 as t=32n-(3.27*2.45)
naw, you want components 'along' or 'into' the ramp.
in your sketch mg is a component of some other vector... I don't know what it is
I got -3.8 m/s^2 (g=9.8) that's the answer your book has, no doubt. so try my way:)
first i did it with no friction o see what would happen and determine my coordnate system. in x2 i solved T=32N-m2a, i put that in x1 and got 32N-m1a-102Nsin(40)=(102N/g)a that turned to 32N-102Nsin(40) = a(10.4+3.37) then through algebra i got a = 2.45
it's not that i'm not trying your way i think were drawing it different but we have the same 102Nsin(40)
when i redrew my FBD for the (b) Find the acceleration of A if A is moving up the incline. nothing changed in x2 so i assumed it would be the same. maybe i have to redo the problem and see if i can get the answer you got.
Thanks Algebraic, i was being lazy and thought the tension would be the same. your right i t was -3.88 m/s^2