# Newton's Law System With Both Frictions

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1. Nov 2, 2015

### Michele Nunes

1. The problem statement, all variables and given/known data
Find the minimum mass that is needed to get the system moving and the acceleration of the system once it starts to move. The coefficient of static friction between the mass and the table is 0.54 and the kinetic friction coefficient is 0.3.

2. Relevant equations
fsmax=(coefficient of static friction)(FN)
fk=(coefficient of kinetic friction)(FN)
FNET of SYS=(msys)(asys)

3. The attempt at a solution
I calculated static friction to be 32.94 N and kinetic friction to be 18.3 N and I know that the normal force and the weight of the first block balance out and I know that the tension in the rope balances out so the only forces I need to take into consideration are friction and the weight of the second block but I don't know what to do with static and kinetic friction, which one do I use and why?

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2. Nov 2, 2015

### BvU

So up to the time the block starts moving, the friction force is $\mu_s F_N$. That helps you to calculate the ?? mass.
When the 6.1 kg moves, you can draw a new free body diagram, but now the friction force is $\mu_k F_N$. That helps you to calculate the acceleration.

 you may assume the minimum mass to get moving is the ?? mass and the changeover from staying at rest and moving is achieved without increasing that mass (a negligible gentle push from a fly bumping its nose or something)

3. Nov 2, 2015

### Staff: Mentor

You also need to draw a free body diagram on the ?? mass, because that mass is accelerating as well, and that affects the tension in the string (i.e., the tension in the string is not equal to the weight of the ?? mass).

4. Nov 2, 2015

### BvU

I sure second that ! Thanks Chet.