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Jusaquikie
Group Title
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.
 one year ago
 one year ago
Jusaquikie Group Title
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.
 one year ago
 one year ago

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Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
post the figure or a diagram.
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
i have T102Nsin(40)19.5=m1a =6.09
 one year ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
block 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.
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
i think i keep confusing my cos and sin
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
hold up. I think I may have confused sin and cos. lol.
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
yeah I switched them, sorry: block on ramp: T 102N*sin(40)  uk*102N*cos(40) = (102N/g)*a
 one year ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
dw:1348867218201:dw
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
how'd you find T?
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
i 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)
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
naw, you want components 'along' or 'into' the ramp.
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
in your sketch mg is a component of some other vector... I don't know what it is
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
I got 3.8 m/s^2 (g=9.8) that's the answer your book has, no doubt. so try my way:)
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
first 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
 one year ago

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

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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