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Jusaquikie
 3 years ago
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.
Jusaquikie
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0post the figure or a diagram.

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0i have T102Nsin(40)19.5=m1a =6.09

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0but i know this is wrong, i think i'm supposed to use the total weight of the system after watching some youtube videos

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0block on ramp: T 102N*cos(40)  uk*102N*sin(40) = (102N/g)*a suspended block: 32NT =(32N/g)*a solve for T T=32N (32N/g)*a plug into first equation and solve for a.

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0i think i keep confusing my cos and sin

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hold up. I think I may have confused sin and cos. lol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah I switched them, sorry: block on ramp: T 102N*sin(40)  uk*102N*cos(40) = (102N/g)*a

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0this isn't different than what i am getting then my numbers are 24N65.6N19.5N divided by 10.4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348867218201:dw

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0i thought we drew that angle like thisdw: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
 3 years ago
Best ResponseYou've already chosen the best response.0naw, you want components 'along' or 'into' the ramp.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in your sketch mg is a component of some other vector... I don't know what it is

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got 3.8 m/s^2 (g=9.8) that's the answer your book has, no doubt. so try my way:)

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0first i did it with no friction o see what would happen and determine my coordnate system. in x2 i solved T=32Nm2a, i put that in x1 and got 32Nm1a102Nsin(40)=(102N/g)a that turned to 32N102Nsin(40) = a(10.4+3.37) then through algebra i got a = 2.45

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0it's not that i'm not trying your way i think were drawing it different but we have the same 102Nsin(40)

Jusaquikie
 3 years ago
Best ResponseYou've already chosen the best response.0when 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
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks Algebraic, i was being lazy and thought the tension would be the same. your right i t was 3.88 m/s^2
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