Finding Tension in a Chain: Understanding Torque and Force Components

In summary: It seems that your confusion is stemming from the fact that the chain is not at an angle of 78 degrees with the rod. The chain is attached at a point that is approximately at the midpoint of the rod.
  • #1
goonking
434
3

Homework Statement



J5T6pCf.png

Homework Equations

The Attempt at a Solution


seems like Torque = r f sin theta should be used here but we don't know r.

Am I suppose to find the x and y components of the tension of the chain?

off topic : if the chain was moved so it would just be above the end of the rod, just like this :
nFy8K4l.png
, would it put less tension on the chain?
 
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  • #2
For variables you don't know, adopt a symbol and go to work. Good chance the thing will drop out towards the end.
 
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  • #3
goonking said:
seems like Torque = r f sin theta should be used here but we don't know r.

You do not always need to know all of the quantities. Why don't you start by simply assuming that the length of the rod is R and do things analytically from there. A good practice is to never try to insert numbers until you have arrived at your final expression.
 
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  • #4
No the chain is in the proper angle to help you calculate the torque of the tension of chain easily ##T_1=31.2r## where r is the length of rod. This torque will be countered by the torque of the weight. Can you find what is the angle ##\phi_1## that the weight makes with the line of rod and then find the torque of weight? You then can see that r is simplified when equating the two torques.
 
  • #5
Problem solving systematics urges you to list the relevant equations and the given/known data.
You do know them already and you do use them already, but still it's a good way to get a grasp on an exercise.

In this case you want ##\Sigma\;\tau = 0## and indeed need ##\tau = \vec r\times\vec F = r F \sin\theta## but this ##\theta## is not the ##\theta## in the drawing that comes with your exercise.

Oh, am I a slow typist. And I need to do some paid work too, so I leave you to the other good helpers...
 
  • #6
Delta² said:
No the chain is in the proper angle to help you calculate the torque of the tension of chain easily ##T_1=31.2r## where r is the length of rod. This torque will be countered by the torque of the weight. Can you find what is the angle ##\phi_1## that the weight makes with the line of rod and then find the torque of weight? You then can see that r is simplified when equating the two torques.
i'm not sure where you mean ϕ1. by weight, do you mean the right end of the rod?
 
  • #7
goonking said:
i'm not sure where you mean ϕ1. by weight, do you mean the right end of the rod?

No i mean the force of weight due to gravity... Can you draw a diagram with all the forces acting on the rod?
 
  • #8
Delta² said:
No i mean the force of weight due to gravity... Can you draw a diagram with all the forces acting on the rod?
should be 78 degrees
 
  • #9
Ok correct, now what is the torque of weight?
 
  • #10
Delta² said:
Ok correct, now what is the torque of weight?
r f sin 78

f = mg

?
 
  • #11
Not exactly. The force of weight is considered to be applied at the c.o.m of the rod. Where is the c.o.m of rod?
 
  • #12
Delta² said:
Not exactly. The force of weight is considered to be applied at the c.o.m of the rod. Where is the c.o.m of rod?
at the middle of the rod
 
  • #13
ok so you have the weight acting on the middle of the rod, at an agle of 78, so what is its torque?
 
  • #14
Delta² said:
ok so you have the weight acting on the middle of the rod, at an agle of 78, so what is its torque?
r/2 x mg x sin 78

?
 
  • #15
yes correct.
 
  • #16
Delta² said:
yes correct.
do we need the torque of the chain? or is that given by the tension in the chain?
 
  • #17
yes you need the torque of the chain. I already said in post #4 what it is. Equate the two torques and solve for m. What do you get?
 
  • #18
Delta² said:
yes you need the torque of the chain. I already said in post #4 what it is. Equate the two torques and solve for m. What do you get?
torque of chain = r 31.2

r 31.2N = r mg sin 78

m = 3.544kg

?
 
  • #19
Didnt we say the torque of weight is (r/2)mgsin(78)
 
  • #20
Delta² said:
Didnt we say the torque of weight is (r/2)mgsin(78)
whoops, yes
 
  • #21
Delta² said:
Didnt we say the torque of weight is (r/2)mgsin(78)

r 31.2N = r/2 mg sin 78

m = 6.5 kg

?
 
  • #22
yes correct. what value did u use for g just asking.
 
  • #23
Delta² said:
yes correct. what value did u use for g just asking.
9.8
 
  • #24
still having a bit of trouble understanding why torque = 31.2 r

isn't the chain at an 78 degree angle too?
 
  • #25
goonking said:
isn't the chain at an 78 degree angle too?

No. The force from the chain is acting perpendicular to the rod and therefore has the full length of the rod as the moment arm.

Edit: It would have a 78 degree angle if it was positioned as in your drawing, but it is not.
 
  • #26
Orodruin said:
No. The force from the chain is acting perpendicular to the rod and therefore has the full length of the rod as the moment arm.

Edit: It would have a 78 degree angle if it was positioned as in your drawing, but it is not.
i see, thank you.
 

1. What is torque and how does it relate to tension?

Torque is a measure of the twisting force applied to an object. It is calculated by multiplying the force applied by the distance from the axis of rotation. Tension is the stretching force applied to an object. The amount of torque applied to an object can affect the amount of tension in that object.

2. How is torque and tension used in engineering and mechanics?

Torque and tension are important concepts in engineering and mechanics because they help determine the strength and stability of structures and machines. Engineers and mechanics use torque and tension calculations to design and build safe and efficient structures and machines.

3. What are the units of torque and tension?

The SI unit for torque is Newton-meters (Nm) and for tension is Newtons (N). In the Imperial system, torque is measured in foot-pounds (ft-lb) and tension is measured in pounds (lb).

4. How do you calculate torque and tension in real-life situations?

To calculate torque, you need to know the force applied and the distance from the axis of rotation. To calculate tension, you need to know the force applied and the area in which the force is applied. In real-life situations, these calculations can be used to determine the strength and stability of structures and machines.

5. What are some examples of torque and tension in everyday life?

Some examples of torque and tension in everyday life include using a wrench to tighten a bolt (applying torque to create tension), a person lifting weights (applying torque to create tension in their muscles), and a bridge supporting the weight of cars (applying tension to withstand the torque of the cars' movement).

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