Question

Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.

A) As mB moves down, determine the magnitude of the acceleration of mA and mB, given θ = 35∘.

B) What smallest value of μk will keep the system from accelerating?

Answer 1

Figure 1: http://session.masteringphysics.com/problemAsset/1057163/4/GIANCOLI.ch05.p030.jpg

Answer 2

@wolfe8

Answer 3

For A, I think you find the F for B going down and that is equal to F of A=ma-N*friction coefficient in the y direction. I think.

Answer 4

I basically had that, only I also have the sine of the angle multiplying the mg.

Answer 5

Draw the free body diagram will ya?! it will clear things out!

Answer 6

I already did that Mashy. What gets me confused are whether or not to multiply by the cosine of the angle or the sine of the angle.

Answer 7

if you draw the free body diagram that confusion will go :P.. can you draw it here?

Answer 8

I'll try to, I kind of stink at drawing on the computer.

Answer 9

Here.. i did it for you
. i forgot to mark the accelerations
but now you can set up equations for each body.. !

Answer 10

m2 gets an acceleration downward , and the same acceleration m1 gets up the incline!

Answer 11

But my master piece! I was almost done.

Answer 12

awwww.. i mmm soo sorryyyY!! :( :( :(

Answer 13

Hang on, small medical emergency. I will be back shortly.

Answer 14

:O is evrything alright? :O.. if i sleep

wolfy you go take it on.. !

Answer 15

Okay, back. My brother got a shard of wood in his foot, but being the macho man he is, is digging it out himself and does not want my help.

Answer 16

ermm.. thats.. ouch!

Answer 17

But thanks Mashy, that is what my FBD looks like on my scratch paper.

Answer 18

No problme.. now all u gotta do is add accelerations and set up equations using newton's second laww!

Answer 19

Okay, so I have:
\[T - m _{A}gsin \theta=\mu _{s}F _{N _{A}}\]

Answer 20

And:
\[T=m _{B}a\]

Answer 21

Is it a in this case or g?

Answer 22

are you doing part A or part B ?

Answer 23

in part A, they are NOT in equilibrium.. so net force is NOT zero..they are accelerating.. thats what i said.. i forgot to mark the accelerations!!

Answer 24

I am on part A.

Answer 25

yea well, then ur equations are wrong, cause ur equations are saying the blocks are in equilibrium
net force = ma not zero .. so can you try again?

Answer 26

Give me a minute, my brother needs help.

Answer 27

alright.. ssup @wolfe8 :D

Answer 28

Okay, back....again. And now I have to get my oars back in the water.

Answer 29

I found in my notes a similar problem, only it does not have the coefficient of friction involved.

Answer 30

Okay, I see where my error was:
\[T-m _{A}gsin \theta-\mu _{S}F _{N _{A}}=m _{A}a\]

Answer 31

And:
\[T=m _{B}g-m _{B}a\]

Answer 32

SOOOOO:
\[m _{B}g-m _{A}gsin \theta -\mu _{S} m _{A}gcos \theta=m _{T}a\]

Answer 33

Woo! I got:
\[a=1.487 m/s ^{2}\]
Which I rounded to \[1.5 m/s ^{2}\]

Answer 34

And I got \[\mu _{K}=0.5206\]
Which I rounded to \[\mu _{K}=0.52\]

Thank you for your help!

Answer 35

your welcome.. OS stopped working for me immediately that time!:-/