Force problem with three masses and two pulleys

In summary, the minimum value of the coefficient of static friction between the block (m3) and the table is calculated using the formula a=((4kg)(9.8m/s^2)-(1kg)(9.8m/s^2)(mu)-(2kg)(9.8m/s^2))/7kg, where the masses do not match the problem statement. However, this approach assumes no friction, whereas in the given problem there is just enough friction to prevent acceleration. To determine the minimum value of the coefficient of static friction, the strength of the frictional force must be calculated.
  • #1
Dayanara Avilez
2
0

Homework Statement


For the apparatus in the figure, what is the minimum value of the coefficient of static friction between the block (m3) and the table that would keep the system at rest if m1 = 0.10 kg, m2 = 0.20 kg, and m3 = 0.30 kg? (Assume ideal conditions for the string and pulleys.)

Homework Equations


F=ma
a= Fnet/mtotal

The Attempt at a Solution


This assignment is extra credit since we haven't been taught how to solve these problems yet. I watched multiple three mass, two pulley videos.They are all solving for acceleration with the mu value. I tried using what I saw to adjust to my situation.

I attempted using the formula a=Fnet/Mtotal
a=((4kg)(9.8m/s^2)-(1kg)(9.8m/s^2)(mu)-(2kg)(9.8m/s^2))/7kg

I plugged in my known values
a=(39.2−(9.8)(μ)−19.6)/7

I then moved my 7 over to the a
7a=39.2−(9.8)(μ)−19.6

I am assuming that I should be able calculate acceleration with my known values since I'm solving for mu but haven't figured out the steps to do so
 
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  • #2
Hi Dayanara Avilez and welcome to PF.

Can you please post the figure? Without it it's hard to check your work.
 
  • #3
Probably be easier to help you if you actually posted the figure you are referencing
 
  • #4
Sorry for the delay! I was figuring out how to attach photos once I saw the replies
phinds said:
Probably be easier to help you if you actually posted the figure you are referencing

upload_2017-11-22_21-9-47.png
 

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  • #5
Dayanara Avilez said:
a=((4kg)(9.8m/s^2)-(1kg)(9.8m/s^2)(mu)-(2kg)(9.8m/s^2))/7kg
These masses do not match your problem statement.
Also, you seem to be calculating the acceleration that would occur without friction. In this question, there is just enough friction to prevent acceleration. So how strong is the frictional force?
 

1. What is a force problem with three masses and two pulleys?

A force problem with three masses and two pulleys is a physics problem that involves calculating the forces acting on three masses connected by a system of two pulleys. The goal is to determine the net force on each mass and the acceleration of the system.

2. How do you approach solving a force problem with three masses and two pulleys?

The first step is to draw a free body diagram to identify all the forces acting on each mass. Then, write out the equations of motion for each mass using Newton's second law (F=ma). Finally, solve the system of equations to find the net force and acceleration of the system.

3. What are the assumptions made in a force problem with three masses and two pulleys?

Some common assumptions include assuming the pulleys are massless and frictionless, and that the strings connecting the masses do not stretch or break. These assumptions allow for a simplified analysis of the problem.

4. How do you account for the direction of forces in a force problem with three masses and two pulleys?

In a force problem with three masses and two pulleys, the direction of forces can be determined by considering the direction of motion of each mass and the direction of the string connecting the masses to the pulleys. Forces acting in the same direction as motion are considered positive, while forces acting in the opposite direction are considered negative.

5. Can you apply the same approach to solve a force problem with different numbers of masses and pulleys?

Yes, the same approach can be applied to solve force problems with any number of masses and pulleys. The key is to draw a clear and accurate free body diagram and write out the equations of motion for each mass. However, as the number of masses and pulleys increases, the calculations can become more complex and time-consuming.

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