What is the largest pulling force before the string breaks?

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Homework Help Overview

The discussion revolves around a physics problem involving three crates being pulled over a smooth horizontal surface at an angle. The crates are connected by strings that can withstand a maximum tension before breaking, and participants are exploring the largest pulling force that can be applied without exceeding this limit.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between tension in the strings and the acceleration of the crates. There are attempts to analyze the forces acting on each crate and the implications of the maximum tension allowed. Questions arise about which string experiences the highest tension and how to determine the maximum allowable acceleration without breaking the strings.

Discussion Status

The discussion is ongoing, with participants providing insights into the forces acting on the crates and the tensions in the strings. Some guidance has been offered regarding the need to analyze the forces on each crate and the implications of Newton's laws. Multiple interpretations of the problem are being explored, particularly regarding the relationship between the tensions in the strings and the applied force.

Contextual Notes

Participants are working under the constraint that the maximum tension each string can support is 38 N. There is also an emphasis on understanding the dynamics of the system rather than jumping to conclusions about the solution.

Johnny1999

Homework Statement



A person pulls three crates over a smooth horizontal floor at an angle of 34 degrees to the horizontal. The crates are connected to each other by identical horizontal strings A and B, each of which can support a maximum tension of 38.0 N before breaking. What is the largest pulling force that can be exerted without breaking either of the strings?
Freed_CollPhysics1e_Figure4-45.jpg


Homework Equations



F = ma,

The Attempt at a Solution


Since the tension for this case is the force, I can find the acceleration by using the max tension 38 / 25 and got 1.52 m/s^2. I'm currently stuck on what to do next
 

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Johnny1999 said:

Homework Statement



A person pulls three crates over a smooth horizontal floor at an angle of 34 degrees to the horizontal. The crates are connected to each other by identical horizontal strings A and B, each of which can support a maximum tension of 38.0 N before breaking. What is the largest pulling force that can be exerted without breaking either of the strings?
View attachment 213658

Homework Equations



F = ma,

The Attempt at a Solution


Since the tension for this case is the force, I can find the acceleration by using the max tension 38 / 25 and got 1.52 m/s^2. I'm currently stuck on what to do next
All the three crates move with the same acceleration, given by the applied force and the masses. In which rope is the tension the highest? Think, rope A has to accelerate the 25 kg mass only, while the tension in rope B has to overcome the force rope A exerts on the mass B and accelerate it with the common acceleration. Draw the FBD for all masses.
 
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ehild said:
All the three crates move with the same acceleration, given by the applied force and the masses. In which rope is the tension the highest? Think, rope A has to accelerate the 25 kg mass only, while the tension in rope B has to overcome the force rope A exerts on the mass B and accelerate it with the common acceleration. Draw the FBD for all masses.
From the FBD I would derive this:
ΣF25 kg = Tension A.
ΣF18 kg = Tension B - Tension A.
ΣF15kg = Tension at angle 34 degrees - Tension B
 
Also, wouldn't Tension A = Tension B because of Newton's 3rd Law and the ropes support only up to 38 N before they break?
 
Johnny1999 said:
From the FBD I would derive this:
ΣF25 kg = Tension A.
ΣF18 kg = Tension B - Tension A.
ΣF15kg = Tension at angle 34 degrees - Tension B
It is the horizontal component of the pulling force.ΣF15kg =Fx - Tension B
No, tension A is not the same as tension B. You have to determine the acceleration and the pulling force so as neither tension A nor tension B is greater than 38 N.
 
ehild said:
It is the horizontal component of the pulling force.ΣF15kg =Fx - Tension B
No, tension A is not the same as tension B. You have to determine the acceleration and the pulling force so as neither tension A nor tension B is greater than 38 N.
So I used the 38 N to find the acceleration of the 25 kg box since the only force acting on it is the tension of rope A and the whole system accelerate at the same rate. I would get 1.52 m/s^2. Is that the right answer?
 
Johnny1999 said:
I used the 38 N to find the acceleration of the 25 kg box since the only force acting on it is the tension of rope A
Ok, but what about rope B? What is the maximum acceleration of the box train before that breaks?
 
haruspex said:
Ok, but what about rope B? What is the maximum acceleration of the box train before that breaks?
How would you figure that out? In order for rope B to break, the pulling force on the 15 kg box in the x direction has to bigger than 38 right?
 
Johnny1999 said:
How would you figure that out? In order for rope B to break, the pulling force on the 15 kg box in the x direction has to bigger than 38 right?
Yes, but consider all the forces on the middle block. Alternatively, treat rope A and the two attached boxes as a system and analyse the forces and acceleration of that.
 
  • #10
haruspex said:
Yes, but consider all the forces on the middle block. Alternatively, treat rope A and the two attached boxes as a system and analyse the forces and acceleration of that.
From FBD I would get this:
M*a (25kg) = 38 N
M*a (18kg) = Tension B - 38 N
 
  • #11
Johnny1999 said:
M*a (25kg) = 38 N
No, don't assume tension A is at max. You need to find out whether B would break first.
 
  • #12
haruspex said:
No, don't assume tension A is at max. You need to find out whether B would break first.
In that case:
a(m1+m2) = Tension B

I used the sum of forces to shorten it to this.
 
  • #13
Johnny1999 said:
In that case:
a(m1+m2) = Tension B

I used the sum of forces to shorten it to this.
Right, so what is the max acceleration allowed by rope B?
 
  • #14
Johnny1999 said:
So I used the 38 N to find the acceleration of the 25 kg box since the only force acting on it is the tension of rope A and the whole system accelerate at the same rate. I would get 1.52 m/s^2. Is that the right answer?
No. Forget the 38 N for the time being. Assume you have Fx horizontal force. What would be the acceleration of all the crates?
 

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