- #1

SakuRERE

- 66

- 5

## Homework Statement

a mass m1 is attached to a second mass m2 by an acme (massless, unstretchable) string. m1 sits on a table with which it has coefficients of static and dynamic friction μs and μk respectively. m2 is hanging over the ends of a table, suspended by the taut string from an acme pulley. at time t=0 both masses are released.

1. what is the minimum mass m2,

_{min}such that the two masses begin to move?

2. If m2= 2m2,

_{min}, determine how fast the two blocks are moving when mass m2 has fallen a height H (assuming the m1 hasn't yet hit the pulley)?

## Homework Equations

Σf=ma

fsmax=μs*fn

vf

^{2}=vi

^{2}+2aΔx[/B]

## The Attempt at a Solution

for a:

I am not sure about how I understand the question, but what comes to my mind when reading ( Begin to move) is that still the system acceleration is zero and that they are asking for the F

_{smax}(the maximum static friction) just before the system starts to move and the friction becomes Kinetic friction of Fk.

so:

for block m1

Σƒ= m*a=0

f

_{smax}=T

μs*Fn= T

**μs*m1g=T**

now for block m2:

Σƒ=m*a=0

**m2g=T**

and by this m2g=μs*m1g so m2=(μs*m1g)/g

and by this m2g=μs*m1g so m2=(μs*m1g)/g

section b)

I guess since the mass of m2 is doubled, that here we will deal with the kinetic friction rather than the static friction.

since both are linked by a rope with the same pulley than they must have the same acceleration, and we should find it.

and after finding it then we will use this equation:

vf

^{2}=vi

^{2}+2aΔx

where vi= 0

so vf=√(2aΔx).

am I right??