Question Pertaining To Newton's Laws

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SUMMARY

This discussion focuses on a physics problem involving Newton's Laws, specifically applied to a system of two boxes connected by a cord over a frictionless pulley. The masses are m1=3.0 kg and m2=1.0 kg, with m2 on an incline of 30°. Participants analyze the forces acting on both masses, derive equations based on Newton's Second Law, and discuss the conditions under which the tension in the cord becomes zero. The critical conclusion is that the cord goes slack when the horizontal force F exceeds the tension T, leading to the condition F = -T.

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  • Understanding of Newton's Second Law of Motion
  • Knowledge of free body diagrams (FBDs)
  • Familiarity with inclined plane physics
  • Basic vector analysis for force components
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Bashyboy
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Homework Statement


Figure 1 shows a box of mass m2=1.0 kg on a frictionless plane inclined at angle θ=30°. It is
connected by a cord of negligible mass to a box of mass m1=3.0kg on a horizontal frictionless
surface. The pulley is frictionless and massless.
(a) Choose the direction of your coordinate system.
(b) Identify all forces acting on m1and m2.
(c) Write down Newton’s 2nd law for m1 and m2.
(d) If the magnitude of horizontal force F is 2.3 N, what is the tension T in the connecting cord?
(e) What is the largest value the magnitude of F may have without the cord becoming slack?


Homework Equations





The Attempt at a Solution



For (a), does question seem to ask me to draw the coordinate system I want to employ, and label the axes?

For (e), I am sort of confused as to how to answer the question. From the diagram, the direction suggests that the question i impossible to answer, for it acts in the direction of the tension force. Wouldn't F have to be zero, because, if it weren't, then it would cause m1 to accelerate even more, under the influence of both T and F, and m1 would push into the string, causing it to bunch up, whereby it won't be taut any longer.
 

Attachments

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Last edited:
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a)Correct- so what do you choose?
A diagram would be nice...
 
I am terribly sorry--I forgot to upload the documents. Hold on, in one moment they shall be uploaded.
 
The vector for T is incorrect.
 
Which tension force is incorrect?
 
Bashyboy said:
For (a), does question seem to ask me to draw the coordinate system I want to employ, and label the axes?
They also want you to choose a direction for your coordinate system, so that you have consistent sign convention in your equations.

For (e), I am sort of confused as to how to answer the question. From the diagram, the direction suggests that the question i impossible to answer, for it acts in the direction of the tension force. Wouldn't F have to be zero, because, if it weren't, then it would cause m1 to accelerate even more, under the influence of both T and F, and m1 would push into the string, causing it to bunch up, whereby it won't be taut any longer.
Hint: What is the tension in the string when it goes slack?
 
Doc Al said:
They also want you to choose a direction for your coordinate system, so that you have consistent sign convention in your equations.
Are my sign conventions not consistent? The coordinate system for m1 is positive going to the right and going upwards; and the coordinate system for m2 is positive when going to the right and upwards.
Doc Al said:
Hint: What is the tension in the string when it goes slack?

Hmm, I suppose it would be T = 0, but I can't justify why.
 
Last edited:
Bashyboy said:
Are my sign conventions not consistent? The coordinate system for m1 is positive going to the right and going upwards; and the coordinate system for m2 is positive when going to the right and upwards.
Sounds good to me. (Your diagram is so faint that it is difficult to read.)

Hmm, I suppose it would be T = 0, but I can't justify why.
For a rope to be taut means that there is tension in it. No tension, it goes slack.
 
So, the string becomes slack when the condition of F = -T is met?

Also, Enigman had said that my tension force was incorrect, which one, is what I ask?
 
  • #10
Bashyboy said:
So, the string becomes slack when the condition of F = -T is met?
No, the string being slack means that T = 0.

Also, Enigman had said that my tension force was incorrect, which one, is what I ask?
Not sure what he meant. Your first attachment is so faint, that I cannot read your equations. Just type them out.
 
  • #11
Doc Al said:
No, the string being slack means that T = 0.

Then how am I to answer question e); of what consequence does the force F have on the tension force? In the second picture the force F is directed to the right, wouldn't any force directed to the right cause the string to go slack? Following this line of reasoning, wouldn't F have to be zero?


Doc Al said:
Not sure what he meant. Your first attachment is so faint, that I cannot read your equations. Just type them out.

I don't have any equations written on either page I uploaded; I think he was referring to the direction which I have assigned to the tension force is wrong.
 
  • #12
Bashyboy said:
So, the string becomes slack when the condition of F = -T is met?
When does the T become zero, what direction does F need to be in?
Also, Enigman had said that my tension force was incorrect, which one, is what I ask?
Nevermind, got confused between the two FBDs.:redface:
 
  • #13
It needs to be opposite to the direction that it is written. Does the string become slack when F < -T?
 
  • #14
Think of it this way- A horse(m2) is pulling a cart (m1)- In what case would the string connecting them will slack given that a car is pushing it?
 
  • #15
If the horse began moving backwards, towards the cart, the string would become slack.
 
  • #16
Yes, but let's assume (m2) Horse doesn't change the direction of motion. Think how much force the car would exert on the cart so that string is slackened- think intuitively.
BTW I'm still not sure if the third FBD is completely correct...gravity components seem to be off...can't make it out properly though...too faint.
 
  • #17
Newton's second law for m2:

\sum F_x = mg \cos (30^{\circ}) - T = m_2a_2

\sum F_y = N - mg \sin(30^{\circ})

Do those appear correct?

--------

I am confused, as you are referring to a car and also a cart.
 
  • #18
Oh, everything is correct except the gravity components. Shouldn't it be cos(600) or sin(300) in Fx? Check m2's first fbd and the second one.
Also won't the tension disappear if the m2 starts accelerating more than the m1?
 
  • #19
Bashyboy said:
For (e)...cause m1 to accelerate even more, under the influence of [STRIKE]both T and[/STRIKE] F, and m1 would push into the string, causing it to bunch up, whereby it won't be taut any longer.

1>If string isn't taut T doesn't act.
2>How much acceleration(/force F) is needed to slacken the string?
3>∑Fx=mgcos(30∘)−T=m2a2
-What about F? where is it?
 
  • #20
Enigman said:
Oh, everything is correct except the gravity components. Shouldn't it be cos(600) or sin(300) in Fx? Check m2's first fbd and the second one.
Also won't the tension disappear if the m2 starts accelerating more than the m1?

Yes, the tension would disappear. The direction of the acceleration of m1 would be towards the right, correct?
 
  • #21
As of now, I am feeling rather uncertain about this entire problem.
 

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