• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Newton's Second Law - Friction

  • Thread starter Marioqwe
  • Start date
1. The problem statement, all variables and given/known data
Two masses, m1 = 3.50kg and m2 = 5.00 kg, are on a frictionless tabletop and mass m3 = 7.60kg is hanging from m1. THE COEFFICIENT OF STATIC AND KINETIC FRICTION BETWEEN m1 AND m2 are 0.60 and 0.50 respectively.

a) What are the acceleration of m1 and m2?



2. Relevant equations

Equation for kinetic and static friction and newton's second law.



3. The attempt at a solution

I found the acceleration for m1 to be 5.2 m/s^2 by doing

a = ((m3)g - Fk) / (m1 + m3)

Now, I don't know what to do in order to find the acceleration of m2.
Do I have to consider the kinetic friction between m2 and m1 and Newton's third law meaning that m1 exerts a force equal to Fk (force due to kinetic friction) on m2?
 
Last edited:
are m1 and m2 connected by string or what?
 
are m1 and m2 connected by string or what?
No no. m1 is resting on top of m2. And m2 is on top of a frictionless table. I forgot to add the part in capital letters to the problem.
 
you can find the friction acting on m1 due to m2
same will act on m2 and cause acceleration in it
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top