What is the largest pulling force before the string breaks?

In summary, the problem involves a person pulling three crates connected by two ropes that can withstand a maximum tension of 38 N. The person is pulling at an angle of 34 degrees to the horizontal. The task is to determine the largest pulling force that can be exerted without breaking either of the ropes. By drawing a free body diagram and considering the forces on each crate, it can be determined that the maximum acceleration allowed by rope B is equal to the tension in rope B divided by the sum of the masses of the two crates connected by it. The acceleration of all three crates can then be calculated using this value and the horizontal force being applied.
  • #1
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
 

Attachments

  • Freed_CollPhysics1e_Figure4-45.jpg
    Freed_CollPhysics1e_Figure4-45.jpg
    14.8 KB · Views: 1,665
Last edited by a moderator:
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes Johnny1999
  • #3
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
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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?
 
  • #8
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?
 
  • #9
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?
 

FAQ: What is the largest pulling force before the string breaks?

What factors affect the strength of a string?

The strength of a string can be affected by several factors, including the type of material it is made of, its thickness, and its length. Other factors such as temperature and tension can also impact the strength of a string.

How is the strength of a string measured?

The strength of a string is typically measured in pounds, kilograms, or newtons. This measurement is usually determined by applying a pulling force to the string until it breaks, and then recording the amount of force that was exerted.

What is the strongest type of string?

The strength of a string can vary depending on its intended use. However, some of the strongest types of string include Kevlar, Spectra, and Dyneema, which are all made from synthetic fibers and are commonly used in ropes and cables that require high strength.

Can a string's strength be improved?

Yes, the strength of a string can be improved by using a thicker and/or stronger material, increasing its length, and applying tension to it. Additionally, proper storage and handling of the string can help maintain its strength over time.

What is the largest pulling force a string can withstand?

The largest pulling force a string can withstand before it breaks will vary depending on the factors mentioned above. However, some of the strongest strings have been known to withstand forces of up to 1000 pounds or more.

Similar threads

Resolving forces: Mass on a string
Replies
4
Views
3K
A rope, a pole, and some tension
Replies
11
Views
1K
Bead on a rod which is being pulled by an ideal string with velocity v
Replies
27
Views
297
Force on a Pulley: Understanding Tension and Tangents
Replies
6
Views
3K
If I pull a pulley, what is the tension in the string around it?
Replies
4
Views
1K
Tension in a string in circular motion
Replies
15
Views
5K
Tension in a string being pulled from both ends
2
Replies
48
Views
74K
Does Tension Increase with Higher Angles in a String?
Replies
2
Views
1K
What is the correct tension equation for a pendulum at rest?
Replies
2
Views
1K
Newton’s Law of Motions: tension forces in a pulley
Replies
4
Views
1K

Hot Threads

Recent Insights

Back
Top