Calculating Acceleration of Hanging Mass System

AI Thread Summary
To calculate the acceleration of the hanging mass system with M1 = 1 kg and M2 = 2 kg, it's essential to analyze the forces acting on each mass separately. For M1, the equation should account for gravitational force and tension, while for M2, only horizontal forces should be considered. The tension in the string acts vertically on M1 and horizontally on M2, necessitating separate treatment of the forces. The discussion emphasizes the importance of correctly identifying the direction of forces and maintaining consistency in the coordinate system. A clearer understanding of the system's layout, particularly the orientation of the string, is crucial for solving the problem.
physicos
Messages
46
Reaction score
1

Homework Statement


A hanging mass M1 =1 kg is attached by a light string that runs over a frictionless pulley to a mass M2=2 kg that is initially at rest on a frictionles table .
What is the magnitude of the acceleration a of M2 ?


Homework Equations


I used for M1 :
Fg1 + T =M1a (with T tension of string)
and for M2 :
N+Fg2+T=M2a

The Attempt at a Solution


NOW I'm stuck , I can't solve it , any help ??
 
Physics news on Phys.org
physicos said:
and for M2 :
N+Fg2+T=M2a
For M2 you have added forces that are perpendicular to each other. Don't do that! Instead, consider horizontal forces separately. (Assuming the table is horizontal, that's the direction of the acceleration.)

Be careful with signs.
 
What I have written are vectors :
for M1 without vectors :
_m1g+T=-m1a
and for M2 :
N+T-m2g=m2a

I still didn't get your point
 
physicos said:
What I have written are vectors :
for M1 without vectors :
_m1g+T=-m1a
Good. (Note that the forces of gravity and tension all act vertically.)

and for M2 :
N+T-m2g=m2a
This combines vertical forces (N, mg) with horizontal forces (T). Don't do that.
 
I thought T was a vertical force too
 
physicos said:
I thought T was a vertical force too

On mass 1 it is a vertical force. On m2 it is pulling from the side. Remember that since your pulley is massless and frictionless, the tension along the string will be completely constant.
 
physicos said:
I thought T was a vertical force too
Not when it acts on M2, which slides along a horizontal table. (A picture would help.)
 
Doc Al said:
Not when it acts on M2, which slides along a horizontal table. (A picture would help.)
The OP doesn't actually state the orientation of the string between pulley and M2. Yes, it's probably meant to be horizontal, but in principle could be anything.
 
so what am I supposed to do ? Cause that's all what is available in the problem statement , there is no picture : Should I work with it as a horizontal force ?
 
  • #10
physicos said:
so what am I supposed to do ? Cause that's all what is available in the problem statement , there is no picture : Should I work with it as a horizontal force ?
Yes.
 
Back
Top