Is the Tension in the String Equal for Both Blocks?

AI Thread Summary
The discussion centers on the tension in a string connecting two blocks with different masses and forces acting on them. It is clarified that the tension must be equal throughout the string, despite the differing forces applied to each block. The equations of motion for both blocks indicate that the acceleration of the system is the same, leading to the conclusion that tensions Ta and Tb must be equal. The confusion arises from the differing magnitudes of the forces acting on each block, but it is emphasized that a massless string cannot have different tensions at different points. Ultimately, understanding the system's acceleration and correcting the equations will allow for solving the tension in the string.
theunloved
Messages
43
Reaction score
1
4.0 kg block A and 6.0kg block B are connected by a string of negligible mass. Force a Fa = 12N acts on block A; force B Fb = 24N acts on block B. What is the tension in the string.

I was able to draw FBD and lists all the forces acting on A and B, but what I don't understand is the tension T acting on A and B.
A:
Fa - Ta = m1 A

B:
Fb - Tb = m2 A

In my understanding, Ta and Tb shouldn't be equal, since we have Fa and Fb exerts on each object and they have different value. What if Ta > Tb, then the cord between A and B should come loose, therefore we'd not have tension at all. But if we don't have have Ta = Tb, then there's no way we can solve the problem...
So my question is: Is it really Ta = Tb and why ?
 
Physics news on Phys.org
The tension is equal, it's just one string after all. What directions are the forces acting? If they are opposite from one another, your equations are close, but one of the applied forces should be negative, with tension adding, while one of the applied forces should be positive, with tension subtracting.

Knowing mass, and knowing the acceleration of the system is equal, and knowing tension is equal, you can solve for tension. You're nearly there, correct your equations, solve both for A, then set them equal to one another, and solve for T. At least that's how I remember doing these.
 
1MileCrash said:
The tension is equal, it's just one string after all. What directions are the forces acting? If they are opposite from one another, your equations are close, but one of the applied forces should be negative, with tension adding, while one of the applied forces should be positive, with tension subtracting.

Knowing mass, and knowing the acceleration of the system is equal, and knowing tension is equal, you can solve for tension. You're nearly there, correct your equations, solve both for A, then set them equal to one another, and solve for T. At least that's how I remember doing these.

Oh, yeah,
Both Fa and Fb have the same direction, going from left to right.
So it should be
Tension in the string
4.0 kg block A and 6.0kg block B are connected by a string of negligible mass. Force a Fa = 12N acts on block A; force B Fb = 24N acts on block B. What is the tension in the string.

I was able to draw FBD and lists all the forces acting on A and B, but what I don't understand is the tension T acting on A and B.

A:
Fa + Ta = m1 A

B:
Fb - Tb = m2 A

I know that its from one string, the tensions should be equal, but, here, we have Fa and Fb. Don't we on the force acting on A and B ? they are different, and they are the forces that make A and B moving. So, if you have Fa > Fb, A will be moving faster than B, and since it's moving faster, it'll catch up with B in a certain time, if that's the case, shouldn't the tensions be different ?
 
A single massless string can not have different tensions at different parts
 
theunloved said:
Oh, yeah,
Both Fa and Fb have the same direction, going from left to right.
So it should be
Tension in the string
4.0 kg block A and 6.0kg block B are connected by a string of negligible mass. Force a Fa = 12N acts on block A; force B Fb = 24N acts on block B. What is the tension in the string.

I was able to draw FBD and lists all the forces acting on A and B, but what I don't understand is the tension T acting on A and B.

A:
Fa + Ta = m1 A

B:
Fb - Tb = m2 A

I know that its from one string, the tensions should be equal, but, here, we have Fa and Fb. Don't we on the force acting on A and B ? they are different, and they are the forces that make A and B moving. So, if you have Fa > Fb, A will be moving faster than B, and since it's moving faster, it'll catch up with B in a certain time, if that's the case, shouldn't the tensions be different ?
If Fa > Fb, solve for T. Strings cannot take negative tensions, they go slack. The blocks accelerate independently from each other. What would be the value of T for that case? As noted,the tension in a massless string must be the same throughout.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top