A community for students.
Here's the question you clicked on:
 0 viewing
Jusaquikie
 4 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
 4 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.

This Question is Closed

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

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

Jusaquikie
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0i think i keep confusing my cos and sin

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

anonymous
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348867218201:dw

Jusaquikie
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0naw, you want components 'along' or 'into' the ramp.

anonymous
 4 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
 4 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
 4 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
 4 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
 4 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
 4 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.