What is the largest pulling force before the string breaks?

[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?

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

 

[ehild]

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.

 

[Johnny1999]

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:
ΣF_{25 kg} = Tension A.
ΣF_{18 kg} = Tension B - Tension A.
ΣF_15kg = Tension at angle 34 degrees - Tension B

 

[Johnny1999]

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?

 

[ehild]

From the FBD I would derive this:
ΣF_{25 kg} = Tension A.
ΣF_{18 kg} = Tension B - Tension A.
ΣF_15kg = Tension at angle 34 degrees - Tension B

It is the horizontal component of the pulling force.ΣF15kg =F_x - 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.

 

[Johnny1999]

It is the horizontal component of the pulling force.ΣF15kg =F_x - 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?

 

[haruspex]

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?

 

[Johnny1999]

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?

 

[haruspex]

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.

 

[Johnny1999]

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

 

[haruspex]

M*a (25kg) = 38 N

No, don't assume tension A is at max. You need to find out whether B would break first.

 

[Johnny1999]

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.

 

[haruspex]

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?

 

[ehild]

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?