Calculating Acceleration of Hanging Mass System

In summary, the problem involves a 1 kg mass connected by a light string to a 2 kg mass on a frictionless table. The question is asking for the magnitude of the acceleration of the 2 kg mass. The solution involves using the equations Fg1 + T = M1a and N + T - M2g = M2a, with T being the tension in the string, N being the normal force, and Fg1 and Fg2 being the forces of gravity on M1 and M2 respectively. When solving, it is important to separate the horizontal and vertical forces and be careful with signs.
  • #1
physicos
46
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 ??
 
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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
I thought T was a vertical force too
 
  • #6
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.
 
  • #7
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.)
 
  • #8
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.
 
  • #9
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.
 

1. How do you calculate the acceleration of a hanging mass system?

To calculate the acceleration of a hanging mass system, you need to use the formula a = (m1 - m2)g / (m1 + m2), where a is the acceleration, m1 is the mass of the object being pulled, m2 is the mass of the hanging object, and g is the acceleration due to gravity (9.8 m/s²).

2. What is the significance of calculating acceleration in a hanging mass system?

Calculating acceleration in a hanging mass system helps us understand how the system is moving and how much force is being applied to the objects. It also allows us to make predictions about the system's future movements.

3. What are the units for acceleration in a hanging mass system?

The units for acceleration in a hanging mass system are meters per second squared (m/s²). This represents the change in velocity over time.

4. How does the mass of the hanging object affect the acceleration in a hanging mass system?

The mass of the hanging object directly affects the acceleration in a hanging mass system. The greater the mass of the hanging object, the smaller the acceleration will be, as there is more weight pulling down on the system.

5. Can the acceleration of a hanging mass system ever be negative?

Yes, the acceleration of a hanging mass system can be negative if the hanging object has a greater mass than the object being pulled. This means that the system is decelerating, or slowing down, rather than accelerating.

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