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Pulley Connecting Mass 1 to Hanging Mass 2

by maximoanimo
Tags: connecting, mass, pulley
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maximoanimo
#1
Sep8-10, 06:53 PM
P: 6
1. The problem statement, all variables and given/known data
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.


2. Relevant equations
F = ma


3. 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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PhanthomJay
#2
Sep8-10, 07:18 PM
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P: 6,038
Quote Quote by maximoanimo View Post
1. The problem statement, all variables and given/known data
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.


2. Relevant equations
F_net = ma


3. 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.
maximoanimo
#3
Sep8-10, 10:13 PM
P: 6
Thanks a lot man. I messed with it for a minute and figured it out.


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