# 3 objects connected by ropes (tension)

• metalmagik
In summary: I'll try again.If you followed the equation, and if you didn't do any 'calculator typos', it should be correct.wow okay I got it wrong...I guess it must've been my fault. I'll try again.In summary, the drawing shows three objects, with m1 = 11.5 kg and m2 = 23.5. They are connected by strings that pass over massless and frictionless pulleys. The objects move, and the coefficient of kinetic friction between the middle object and the surface of the table is 0.100.
metalmagik
The drawing shows three objects, with m1 = 11.5 kg and m2 = 23.5. They are connected by strings that pass over massless and frictionless pulleys. The objects move, and the coefficient of kinetic friction between the middle object and the surface of the table is 0.100.

(a) What is the acceleration of the three objects?

(b) Find the tension in each of the two strings.

I made FBDs for these three blocks...and what I got ended up being the acceleration for the middle block, which came out to be $1.7 m/s^2$. How do I find the acceleration of ALL three objects?? Any help is greatly appreciated, this is for a webassign and I would really like to get this correct.

metalmagik said:
I made FBDs for these three blocks...and what I got ended up being the acceleration for the middle block, which came out to be $1.7 m/s^2$. How do I find the acceleration of ALL three objects?? Any help is greatly appreciated, this is for a webassign and I would really like to get this correct.
HINT: All three block accelerate at the same rate.

so if I find the mass of ONE block, I've found the mass at which all three blocks accelerate? or must I add them up?

I do not understand how to achieve the acceleration when you don't have the tension forces. You need those to achieve Fnet do you not?

metalmagik said:
I do not understand how to achieve the acceleration when you don't have the tension forces. You need those to achieve Fnet do you not?

All you have to do is slow down. Take the mass m2 and draw a FBD for it. As Hootenanny said, all masses accelerate at the same rate. Apply Newton's second law to mass m2. Then do the same thing for the mass in the middle. And then for the left one. Look at the equations you got and try to solve for a.

Fnet = ma...so...for the right one what is the Fnet? I still don't understand how one does this without the Force of tension. I'm very confused.

metalmagik said:
Fnet = ma...so...for the right one what is the Fnet? I still don't understand how one does this without the Force of tension. I'm very confused.

OK, let's take the right mass, m2. The forces acting on it are the gravitational force G = m2*g and the tension in the right rope, let's call it T2. So, tkat makes Fnet = T2 - G = T2 - m2*g. When you use Newton's second law, you have Fnet = m2*a, i.e. T2 - m2*g = m2*a. Now do the same thing for the other masses.

Ok so for the center one I have $T1-F+T2 = Mc*a$

and for the left one I have $T1 - m1g=m1a$

is this correct?

metalmagik said:
Ok so for the center one I have $T1-F+T2 = Mc*a$

and for the left one I have $T1 - m1g=m1a$

is this correct?

Correct, but you made a mistake for the central mass. T1 and T2 are pointing in opposite directions. So, it should be -T1 - F + T2 = Mc*a.

oh I see. How can I now calculate acceleration from this?

metalmagik said:
oh I see. How can I now calculate acceleration from this?

You have three equations now, right? Well, use the equations for the right and left mass, and express them in a form T2 = ... and T1 = ... Then plug T1 and T2 into the equation for the central mass and you'll be able to calculate the acceleration a out of this equation directly.

I have -(m1g+m1a)-Ff+(m2g+m2a) = ma

Is it possible for me to set the accelerations equal to 1 making them insignificant in the equation?

metalmagik said:
I have -(m1g+m1a)-Ff+(m2g+m2a) = ma

Is it possible for me to set the accelerations equal to 1 making them insignificant in the equation?

I don't understand what you're trying to ask. You now know everything in the equation, just solve for a.

Oh I see, I got .871 $m/s^2$

Is this correct? I am going CRAZY with all of this haha

metalmagik said:
Oh I see, I got .871 $m/s^2$

Is this correct? I am going CRAZY with all of this haha

If you followed the equation, and if you didn't do any 'calculator typos', it should be correct.

wow okay I got it wrong...I guess it must've been my fault.

## 1. What is tension in a system of 3 objects connected by ropes?

Tension is the force that is transmitted through a rope when it is pulled or stretched. In a system of 3 objects connected by ropes, the tension will be the same throughout all of the ropes as long as they are all connected to each other.

## 2. How is tension calculated in a system of 3 objects connected by ropes?

Tension is calculated by using Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. In a system of 3 objects connected by ropes, the tension can be calculated by considering the forces acting on each object and setting them equal to each other.

## 3. What happens to the tension in a system of 3 objects connected by ropes if one of the ropes breaks?

If one of the ropes breaks, the tension in the remaining ropes will increase. This is because the total force being applied to the system is now distributed between fewer ropes, resulting in a higher tension in each individual rope.

## 4. Can the tension in a system of 3 objects connected by ropes ever be zero?

No, the tension in a system of 3 objects connected by ropes cannot be zero. This is because in order for the system to remain in equilibrium, the tension in each rope must be equal and non-zero. If the tension in one of the ropes were to be zero, the system would become unbalanced and one of the objects would begin to move.

## 5. How does the angle of the ropes affect the tension in a system of 3 objects connected by ropes?

The angle of the ropes does not affect the tension in a system of 3 objects connected by ropes. As long as the ropes remain taut and connected to each other, the tension will be the same regardless of the angle at which they are pulled.

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