Rope tension

  • Thread starter missrikku
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  • #1
Hello again! I just wanted to make sure that I am understanding tension force. Okay, there are 3 masses (A, B, C) and they are connected by some massless and unstrectchable rope and they are pulled with a force of T3 = # N and they are pulled on a horizontal and frictionless table:


Am I correct in thinking that T1 = T2 = T3 = # N (whatever value # is)? I think that the cord connected to C with force of T3 also pulls the other bodies, A and B, with the same magnitude of T3.

Also, since T3 is the magnitude of the force pulling all 3 of these bodies, can I combine the masses of the bodies and just use the value of T3 to figure out the acceleration of the system?

Ma + Mb + Mc = Msys
T1 = T2 = T3 = Tsys

F = ma
F = T = Msys(a)
a = T/Msys


Answers and Replies

  • #2
Nope, T1, T2, and T3 are not equal... It's show time for Newton's third law!

Newton's third law: If a body A excerts a force F to body B, then body B must exert a force F, opposite in direction, to body A. Examples:

1) If an electron excerts an attractive force to a proton, then we know from Newton's third law that the electron must also feel an attractive force excerted by the proton.

2) If I push the wall with a force of Z Newtons, the the wall also pushes me with the force of Z Newtons (heck, that's why my hands can feel the existence of the wall).

3) We know that the Earth excerts a gravitational force on a falling apple, so we can conclude (from Newton's third law) that the apple excerts the same amount of gravitational force on the Earth (If the Earth doesn't seem to move towards the apple, that's because Earth is very very heavy).

After you have faith in Newton's third law, let's go to your system... We'll start with A. Obviously A accelerates to the right. Let's say a force F1 causes this acceleration. Let's draw the force diagram for A:

A ---> (F1)

Who can probably give rise to this force? It must be B since A is connected to B by a rope. But by Newton's third law, If B excerts a force to A, then A must also exert a force to B. The force will be equal in magnitude, but in the opposite direction. Let's draw the force diagram for B:

(F1) <--- B

By Newton's third law, F1 must exist. But is that diagram complete? Wait a moment! If there is only F1 on B, then B must be accelerating to the left! That is not the case, so there must be another force that makes B accelerates to the right. The force must be greater than F1 (do you know why?). Let's make the correct force diagram:

(F1) <--- B ------> (F2)

Now, who can possibly exert F2? It must be C since C is connected to B by a rope. Without repeating the previous explanation, the force diagram for C is like this:

(F2) <------ C ---------> (F3)

Who can exert F3? It's no other than than the person/car/whatever pulling the system. Note that in our system, the rope is massless so it only serve to transfer force.

By your definition, F3 is # N (it would be a lot better if you use letters like a, b, and c instead of # for variables). And by the previous discussions, we know that
F2 is less than F3
F1 is less than F2

Note that we don't draw the gravitational and normal force in our diagram. Their sum is zero so it won't affect the motion of our masses.

Some equations to note is:

aC = (F3 - F2) / mC
aB = (F2 - F1) / mB
aA = F1 / mA

And aA = aB = aC (all the masses have the same acceleration!)

If we call the acceleration a, then it is also the case that:

a = F3 / (mA + mB + mC)

That is the case, since we can regard the three masses as one object (like you can regard A as an object. if A is a box, then A is actually composed of smaller mass components, right?). If we regard A, B, and C as an object O, then it's force diagram is simple like this:

O ---------> (F3)

Because the object O has the mass (mA + mB + mC), we can derive the previous equation.

Is my explanation helpful?

A question: what are the tension in the ropes?

May the force be with you (no pun intended)...
  • #3
Science Advisor
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Wow! More than you ever wanted to know about Newton's third law :smile:.

missrikku : Here is my simple minded way of doing that problem:

A, B, and C are all pulled by force T3. That means that force T3 is acting on a total mass of (A+B+C) and, so, is accelerating them with acceleration (f/m) T3/(A+B+C). Force T2 is acting only on A and B. Since their mass is (A+B) but they are accelerating at T3/(A+B+C), force T2 must be (ma) T3(A+B)/(A+B+C). Force T1 is acting only on A. Since it is accelerating at T3/(A+B+C), T1 must be T3A/(A+B+C).

Another way to do this is to find that the joint acceleration of A,B,C is T3/(A+B+C) as above. Since C alone is accelerating with that acceleration, the NET force on it must be T3C/(A+B+C). We already know that T3 is pulling it one way. The force back the other way (T2) must be T3- T3C/(A+B+C)= (T3(A+B+C)- T3C)/(A+B+C)= T3(A+B)/(A+B+C) just as before.

You calculate T1 even easier: it is just enough to accelerate A at T3/(A+B+C).
  • #4
wow! Thanks much! I think I asked about this before, but do you know of any books I could possibly check out at a library to help me understand physics better? I would like to practice more problems. Thank you!

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