Pulley Connecting Mass 1 to Hanging Mass 2

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SUMMARY

The discussion focuses on calculating the acceleration of a system involving two masses connected by a cord over a frictionless pulley. The first mass (M1) is 7.01 kg on a table, while the second mass (M2) is 2.35 kg hanging in the air. The correct approach involves recognizing that the tension in the cord affects both masses, leading to the equations of motion for each mass. The final acceleration is determined to be 3.29 m/s², but the initial calculation was incorrect due to misunderstanding the forces acting on M2.

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  • Understanding of Newton's Second Law (F = ma)
  • Basic principles of tension in a pulley system
  • Knowledge of forces acting on connected masses
  • Concept of frictionless surfaces in physics
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  • Explore examples of frictionless motion in physics problems
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Homework Statement


A block with mass 7.01kg rests on a frictionless table. The mass is connected by a cord, passing over a pulley to another block with mass 2.35kg that hangs in the air. Assume the cord to be massless and unstretchable and the pulley to have no friction or rotational inertia.

Find the acceleration of the first block.


Homework Equations


F = ma


The Attempt at a Solution


Call the block with mass 7.01kg M1 and the one with 2.35kg M2. The forces acting upon M1 are the force of gravity, the normal force, and the horizontal force of movement (there is no force of friction since it is said that the table is frictionless). Therefore, the net force on M1 is simply the horizontal force of movement.
The forces acting upon M2 are the force of gravity, the normal force, and the force of the string. The first two cancel, leaving a net force equal to that exerted by the string. Since M1 and M2 are connected by the cord, they exert an equal force on one another, and therefore

F(m1) = F(m2)
(7.01kg)a = (2.35kg)(9.8)
a = 3.29 m/s^2

However, I entered this answer and it said it was incorrect. Is there an error in my thinking, or in my calculation?

Thanks so much for the help!
 
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maximoanimo said:

Homework Statement


A block with mass 7.01kg rests on a frictionless table. The mass is connected by a cord, passing over a pulley to another block with mass 2.35kg that hangs in the air. Assume the cord to be massless and unstretchable and the pulley to have no friction or rotational inertia.

Find the acceleration of the first block.


Homework Equations


F_net[/color] = ma


The Attempt at a Solution


Call the block with mass 7.01kg M1 and the one with 2.35kg M2. The forces acting upon M1 are the force of gravity, the normal force, and the horizontal force of movement (there is no force of friction since it is said that the table is frictionless). Therefore, the net force on M1 is simply the horizontal force of movement.
correct, but you mean the tension force which causes the movement, not the 'force of movemnt'
The forces acting upon M2 are the force of gravity-
yes
the normal force,
What normal force? It's hanging in the air
and the force of the string
yes.
The first two cancel, leaving a net force equal to that exerted by the string.
no
Since M1 and M2 are connected by the cord, they exert an equal force on one another, and therefore

F(m1) = F(m2)
(7.01kg)a = (2.35kg)(9.8)
a = 3.29 m/s^2

However, I entered this answer and it said it was incorrect. Is there an error in my thinking, or in my calculation?

Thanks so much for the help!
The tensionns on each mass are equal, but the net force on each is not. They both must have the same acceleration, a. Examine the forces on each mass and solve 2 equations with 2 unknowns.
 
Thanks a lot man. I messed with it for a minute and figured it out.
 

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