Calculating Tension in a Multi-Duck System

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To calculate tension in a multi-duck system, first determine the total mass and acceleration of the system using the formula F = ma. In the example provided, with two ducks of 5 kg and 2 kg being pulled by a 10N force, the acceleration is calculated to be 1.4 m/s². The tension in the rope can be found by multiplying the acceleration by the mass of the duck being considered, such as the 2 kg duck. When an additional 1 kg duck is added, the tension calculations will depend on the mass of the ducks being pulled and their respective positions in the system. Understanding the relationship between mass, acceleration, and tension is crucial for solving these types of physics problems.
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hi i am in grade 11 university physics, and i have a test coming up that requires me to calculate tension. Unfortunately i have no clue how to do this. Can someone help ?
 
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well it depends on the question you are given. But I'd say (at a random guess) that you need to resolve all the forces in one direction and put that equal to ma, then try to solve from there.
 
ok well my question states that, two toy ducks are attached by a string. the front duck has a mass of 5 kg, and the back duck is 2.0 kg. they are bing pulled wiith a force of 10N and there is no friction.
calculate the tension of the rope after u have calculated the acceleration.

the acceleration of the ducks is 1.4m/s2
and i drew an FBD like soo(2.0kg)DUCk2------(5.0kg)DUCK1-------->10N

i still don't kno where to go from here?
 
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Homework .. ehh

First u have to calculate the accelaration of the system:

F=m(total)*a => a=F/m(total)

m(total)=m1+m2=7 kg

the Tension is equal to the inertial force that the second weight (2kg) exerts. so:

T(ension)=F=a*m2

a-acceleration
m2-second duck 2 kg
 
thanks a lot, that helped me tremendously. :)

i get it now, but there is another part where it says that another duck(1 kg) is added, and it is put at the front of the row.

DUCK2(2kg)------DUCK1(5kg)------DUCK3(1kg)----->(10N)

Now it asks for the tension connecting ducks 2(2kg) and 3(1kg)and then the tension of ducks 1(5kg) and ducks 2(2kg) . would the tension still be equal to the inertial force of the last duck(2kg)? or would it now be equal to the inertial force of the middle duck(5kg)?
 
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mIsix said:
thanks a lot, that helped me tremendously. :)

i get it now, but there is another part where it says that another duck(1 kg) is added, and it is put at the front of the row.

DUCK2(2kg)------DUCK1(5kg)------DUCK3(1kg)----->(10N)

Now it asks for the tension connecting ducks 2(2kg) and 3(1kg)and then the tension of ducks 1(5kg) and ducks 2(2kg) . would the tension still be equal to the inertial force of the last duck(2kg)? or would it now be equal to the inertial force of the middle duck(5kg)?

In all these questions, once you have determined the acceleration for the group, then the tension (or force in between if it is pushing it) will necessarily be the acceleration times all the mass that is being pulled (or pushed as the case may be.)
 
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