Incline with Pulley, find the mass of one block.

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SUMMARY

The problem involves a 20.0 kg block on a frictionless inclined plane at a 30.0-degree angle, connected via a cord over a frictionless pulley to a second block. To achieve an acceleration of 2.00 m/s², the mass of the second block must be calculated. The correct mass of the second block is determined to be 17.7 kg, based on the net force equations and gravitational forces acting on both blocks. The calculations involve isolating forces and applying Newton's second law.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of gravitational force calculations (Fg = mg)
  • Familiarity with inclined plane physics
  • Basic algebra for solving equations
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  • Study the dynamics of systems involving pulleys and inclined planes
  • Learn about net force calculations in multi-body systems
  • Explore the effects of friction on inclined planes
  • Practice solving problems involving acceleration and mass relationships
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Nirupt
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Homework Statement



A 20.0 kg block rests on a frictionless inclined plane of slope angle 30.0 degrees. A light cord attached to the block passes over a frictionlesss pulley at the top of the plane and is attached to a second block. What must be mass of the second block if the system is to be accelerating up and to the right at 2.00 m/s2?

https://dist-ed.waketech.edu/course...63858134f05bd87414264516909/InclineAtwood.png

Is a link to the image.

I found in class that the answer is 17.7 kg.. however I am stumped on where the numbers should go.



Homework Equations





The Attempt at a Solution


http://hyperphysics.phy-astr.gsu.edu/hbase/incpl.html#c1 is the link where I use my formulas

Well I do know that if I isolate the mass on the incline

ƩFnetx = -m1g * sin(30°) + Fτ = 2*20

I got 2*20 because of ma, and I know for m1g I substitute (20*9.8) I also know that acceleration is going up the incline, and if I were to remove the pulley and put them on the x-axis, it would be going to the right which is positive, therefore, the acceleration is positive.

Solving for that I get, Fτ = 138N which I know to be true

Now.. isolating m2 I get
ƩFnetx = -Fτ + m2g = 2m

however I have two masses??

In a previous problem when I had to find the mass of m2 if I wanted the objects to be at rest or constant velocity (so I assumed F=ma, but the sum would be 0). I ended up finding the weight of M2 being 98N.. would I plug that in for m2g? I guess not considering it doesn't give me the answer.. but I just wanted some feedback on this.
 
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Nirupt said:

Homework Statement



A 20.0 kg block rests on a frictionless inclined plane of slope angle 30.0 degrees. A light cord attached to the block passes over a frictionlesss pulley at the top of the plane and is attached to a second block. What must be mass of the second block if the system is to be accelerating up and to the right at 2.00 m/s2?

https://dist-ed.waketech.edu/course...63858134f05bd87414264516909/InclineAtwood.png

Is a link to the image.

I found in class that the answer is 17.7 kg.. however I am stumped on where the numbers should go.



Homework Equations





The Attempt at a Solution


http://hyperphysics.phy-astr.gsu.edu/hbase/incpl.html#c1 is the link where I use my formulas

Well I do know that if I isolate the mass on the incline

ƩFnetx = -m1g * sin(30°) + Fτ = 2*20

I got 2*20 because of ma, and I know for m1g I substitute (20*9.8) I also know that acceleration is going up the incline, and if I were to remove the pulley and put them on the x-axis, it would be going to the right which is positive, therefore, the acceleration is positive.

Solving for that I get, Fτ = 138N which I know to be true

Now.. isolating m2 I get
ƩFnetx = -Fτ + m2g = 2m

however I have two masses??

In a previous problem when I had to find the mass of m2 if I wanted the objects to be at rest or constant velocity (so I assumed F=ma, but the sum would be 0). I ended up finding the weight of M2 being 98N.. would I plug that in for m2g? I guess not considering it doesn't give me the answer.. but I just wanted some feedback on this.
Your last equation should read ƩFnetx = -Fτ + (m_2)g = 2(m_2). Now do the algebra correctly to solve for m_2!
 
So I write the Equation..

m_2 = -138 + m_2g
--------------------
2

However, I still scratch my head at this, and I'm not sure if substituting would work either. Sorry if my algebra skills seem to be lacking... it is 1:37 a.m. where I am at currently.
 
Bring all of the mass dependent quantities to one side and factor it out. Divide the -Ft by accel - grav. You should end up with m_2 = (-138.1)/(2-9.81)
 

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