How do I find the acceleration of a mass on a table?

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SUMMARY

The acceleration of a mass on a table can be calculated by considering the resultant forces acting on the system. In this discussion, the resultant force was determined to be 15.2 N after accounting for friction, leading to an initial incorrect acceleration calculation of 10.1 m/s². The correct approach involves using the dynamic coefficient of friction and treating the entire system as a single mass, resulting in an accurate acceleration of 1.9 m/s². This method emphasizes the importance of force balance and the correct application of friction coefficients in dynamics problems.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Knowledge of static and dynamic friction coefficients
  • Ability to perform force balance calculations
  • Familiarity with basic kinematics and acceleration concepts
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  • Study the application of Newton's Second Law in multi-body systems
  • Learn about the differences between static and dynamic friction
  • Explore tension in strings and its effect on acceleration
  • Practice solving problems involving force balance in accelerating systems
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Students studying physics, particularly those focusing on dynamics and mechanics, as well as educators seeking to enhance their teaching methods in force and motion concepts.

kasse
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[SOLVED] acceleration of mass on table

Homework Statement



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The Attempt at a Solution



The resultant force of the two hanging mases is 19.6 N. This means that just after the masses are released, the friction is 0.3*mg = 4.41 N, so that the total resultant force is 15.2 N.

The acceleration is then 15.2/1.5 = 10.1 m/s^2

Why am I wrong?
 
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First of all, The coefficient of friction you have is for static friction.
Secondly, you must account for the entire system's mass since they are all accelerating.
 
You got an answer larger than 9.8m/s^2. That should be sending up alarms. No matter how heavy m3 is it's maximum acceleration down is 9.8m/s^2. Now try and do the problem right. There is a tension T1 and T2 in each of the two strings. You have to do a force balance for each mass and then set all of the accelerations equal.
 
I solved it in the same way, replacing the static coefficient with the dynamic, and replacing m2 with (m1+2+3). 1.9m/s^2. I don't know if my method is right, but it gave me the correct answer.
 
kasse said:
I solved it in the same way, replacing the static coefficient with the dynamic, and replacing m2 with (m1+2+3). 1.9m/s^2. I don't know if my method is right, but it gave me the correct answer.

That works. As Texag said, since everything is accelerating at the same rate, you can treat the system as one large mass and just add up the external forces. This does save you the step of dealing with the internal tensions.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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