Newton's Law System With Both Frictions

AI Thread Summary
To determine the minimum mass required to initiate movement in the system, static friction must be calculated using the coefficient of static friction (0.54), resulting in a maximum static friction force of 32.94 N. Once the system begins to move, kinetic friction applies, calculated with the coefficient of kinetic friction (0.3), yielding a friction force of 18.3 N. The normal force and weight of the first block balance, while the tension in the rope must be considered, as it changes when the mass accelerates. A free body diagram for both masses is essential to accurately analyze forces and tensions involved. Understanding the transition from static to kinetic friction is crucial for solving the problem effectively.
Michele Nunes
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Homework Statement


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.

Homework Equations


fsmax=(coefficient of static friction)(FN)
fk=(coefficient of kinetic friction)(FN)
FNET of SYS=(msys)(asys)

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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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.

[edit] 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)
 
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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).
 
I sure second that ! Thanks Chet.
 
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