What is the Force Exerted by the Ground on Object 1 in an Atwood Machine?

In summary, the conversation discussed the calculation of the force that the ground exerts on object 1 in different situations where object 2 has different weights. The correct approach was to use the equation w1 - w2 = Fn, which resulted in the ground exerting a force of 47 Newton's in situation (a) and 17 Newton's in situation (b). In situation (c), the ground would exert a force of 0 Newton's, causing object 1 to be "in the air."
  • #1
antonisz
27
0
1. A 77.00 Nt object (denoted as object 1) rests on the ground. A light cord is connected to this object which runs vertically upward over a light frictionless pulley and is attached to another object denoted as object 2.

a) Calculate the force that the ground exerts on object 1 if object 2 is 30 N.

b) Calculate the force that the ground exerts on object 1 if object 2 is 60 N.

c) Calculate the force that the ground exerts on object 1 if object 2 is 90 N.



http://imgur.com/cVEQW07 http://imgur.com/cVEQW07

g (m1 - m2) / (m1) + m2 = ay I drew FBD's for both of the weights and solved for the acceleration for situation a. I then used the equation w - t = ma and solved for the tension value, I then subtracted the tension value from the original weight of object 1, and I got 33.88 N.

I feel like this is the wrong way of solving the problem because once you get to situation c, you would have a negative force which can't be possible.
 
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  • #2
antonisz said:
1. A 77.00 Nt object (denoted as object 1) rests on the ground. A light cord is connected to this object which runs vertically upward over a light frictionless pulley and is attached to another object denoted as object 2.

a) Calculate the force that the ground exerts on object 1 if object 2 is 30 N.

b) Calculate the force that the ground exerts on object 1 if object 2 is 60 N.

c) Calculate the force that the ground exerts on object 1 if object 2 is 90 N.



http://imgur.com/cVEQW07 http://imgur.com/cVEQW07

g (m1 - m2) / (m1) + m2 = ay


I drew FBD's for both of the weights and solved for the acceleration for situation a. I then used the equation w - t = ma and solved for the tension value, I then subtracted the tension value from the original weight of object 1, and I got 33.88 N.

I feel like this is the wrong way of solving the problem because once you get to situation c, you would have a negative force which can't be possible.
That seems to be a correct result for part c. -- but ...

If the ground can't produce a negative (downward) force on m1, then what do you suppose happens to the system?

attachment.png
 

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  • #3
SammyS said:
That seems to be a correct result for part c. -- but ...

If the ground can't produce a negative (downward) force on m1, then what do you suppose happens to the system?

attachment.php?attachmentid=71007&stc=1&d=1404148813.png

The system would change where object 1 would be "in the air" correct?
 
  • #4
antonisz said:
The system would change where object 1 would be "in the air" correct?

Yes, so the ground will exert what force on m1?
 
  • #5
SammyS said:
Yes, so the ground will exert what force on m1?

It would have to be 0 Newton's. If I'm correct about that, then I already assumed that, however I rethought my approach on part a and b.

I redid the FBD's for (a) and (b) and added the two equations for Newton's second law, and the tensions canceled each other out, leaving me with w1 - w2 = Fn. From there I calculated that in situation (a), the force that the ground exerts is 47 Newton's, and in situation (b) it would be 17 Newton's.
 
Last edited:
  • #6
antonisz said:
It would have to be 0 Newton's. If I'm correct about that, then I already assumed that, however I rethought my approach on part a and b.

I redid the FBD's for (a) and (b) and added the two equations for Newton's second law, and the tensions canceled each other out, leaving me with w1 - w2 = Fn. From there I calculated that in situation (a), the force that the ground exerts is 47 Newton's, and in situation (b) it would be 17 Newton's.
That all looks good!
 
  • #7
SammyS said:
That all looks good!

Thank you so much for the help!
 

Related to What is the Force Exerted by the Ground on Object 1 in an Atwood Machine?

What is an Atwood Machine problem?

An Atwood Machine problem is a physics problem that involves two masses connected by a pulley and a string. The problem typically asks for the acceleration of the system, the tension in the string, or the difference in the masses.

What are the basic principles behind solving an Atwood Machine problem?

The basic principles used to solve Atwood Machine problems are Newton's Second Law of Motion, the law of conservation of energy, and the concept of torque. These principles are applied to each mass in the system to determine the acceleration, tension, and other variables.

How do you draw a free-body diagram for an Atwood Machine problem?

To draw a free-body diagram for an Atwood Machine problem, you must first identify the two masses and the pulley. Then, draw arrows to represent the forces acting on each mass, including weight, tension in the string, and normal force. The direction of the arrows should correspond to the direction of the force.

What are some common mistakes when solving Atwood Machine problems?

Some common mistakes when solving Atwood Machine problems include not properly labeling forces on the free-body diagram, not considering the direction of the acceleration, and not setting up the equations of motion correctly. It is also important to pay attention to the signs of the variables, such as positive or negative acceleration and tension.

Are there any real-world applications of the Atwood Machine problem?

Yes, the Atwood Machine problem has real-world applications in fields such as engineering and physics. It can be used to analyze the motion of elevators, cranes, and other systems that involve pulleys and masses. It is also used in the design of mechanical systems and machines.

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