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- Jusaquikie

Body A weighs 102 N, and body B weighs 32 N. The coefficients of friction between A and the incline are μs = 0.56 and μk = 0.25. Angle θ is 40°. Let the positive direction of an x axis be up the incline.
(b) Find the acceleration of A if A is moving up the incline.

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- Jusaquikie

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- anonymous

post the figure or a diagram.

- Jusaquikie

- Jusaquikie

i have T-102Nsin(40)-19.5=m1a =-6.09

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- Jusaquikie

but i know this is wrong, i think i'm supposed to use the total weight of the system after watching some youtube videos

- anonymous

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.

- Jusaquikie

i think i keep confusing my cos and sin

- anonymous

hold up. I think I may have confused sin and cos. lol.

- anonymous

yeah I switched them, sorry:
block on ramp:
T -102N*sin(40) - uk*102N*cos(40) = (102N/g)*a

- Jusaquikie

this isn't different than what i am getting then
my numbers are 24N-65.6N-19.5N divided by 10.4

- Jusaquikie

for x1

- anonymous

|dw:1348867218201:dw|

- anonymous

how'd you find T?

- Jusaquikie

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)

- anonymous

naw, you want components 'along' or 'into' the ramp.

- anonymous

in your sketch mg is a component of some other vector... I don't know what it is

- anonymous

I got -3.8 m/s^2
(g=9.8)
that's the answer your book has, no doubt. so try my way:)

- Jusaquikie

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

- Jusaquikie

it's not that i'm not trying your way i think were drawing it different but we have the same 102Nsin(40)

- Jusaquikie

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.

- Jusaquikie

Thanks Algebraic, i was being lazy and thought the tension would be the same. your right i t was -3.88 m/s^2

- anonymous

cool!

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