Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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.

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

post the figure or a diagram.
1 Attachment
i have T-102Nsin(40)-19.5=m1a =-6.09

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
for x1
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question