Finding Acceleration of Blocks Connected by String

[ƒ(x)]

Problem: I've been assigned a problem that involves a block on a table that is attached by a string to a block hanging over the edge. They are of different masses, and I am given coefficients for both static and kinetic friction. The former is .50, and the latter is .30. I have to find the acceleration of the system if it is released from rest.

My problem: ok, so do I need to use the coefficient of kinetic friction at all? I do not think so...that's pretty much my question.

 

[ƒ(x)]

I got a negative answer..

 

[c7ng23]

Can you show what equations you used? Kinetic friction should be used for surfaces moving relative to each other.

 

[tiny-tim]

Hi ƒ(x)!

Do good ol' Newton's second law on each block separately, plus the fact that their accelerations must be the same (because the string length is constant).

What do you get? ​

 

[ƒ(x)]

m1 (on table) = 10 kg
m2 = 4 kg

.5*Fn = Fs
.5*10*9.8 = 49 N = Fs

Fnet = Fx - Fs = 4*9.8 - 49 = -9.8 N

ma = -9.8
a = -9.8/m = -9.8/(4+10) = -.7 m/s/s

 

[tiny-tim]

m1 (on table) = 10 kg
m2 = 4 kg

ohh! you didn't give the masses before

no wonder you got a negative answer

.5*Fn = Fs
.5*10*9.8 = 49 N = Fs

Fnet = Fx - Fs = 4*9.8 - 49 = -9.8 N

ma = -9.8
a = -9.8/m = -9.8/(4+10) = -.7 m/s/s

That doesn't make sense … how can the mass be accelerating upward?

What does it mean if the weight is less than the µ_sN ?

 

[ƒ(x)]

I'm guessing it means that the system isn't moving.

 

[tiny-tim]

I'm guessing it means that the system isn't moving.

(why guessing? )

That's right! …

if the system is released from rest, it will never move (even though if it was given a little nudge, the low µ_k would enable it to keep accelerating).