How Do I Start Solving Part C of the Physics Problem?

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Part C of the physics problem can be approached by resolving the tension into components parallel and perpendicular to line AB. The angle ABF is crucial for determining the force's contribution to the moment about point A. The component of force parallel to AB does not create a moment, so focus on the perpendicular component, which requires knowing the distance AB. The calculated force should be based on the moment equation, and correcting any errors in the distance or angle will lead to the correct minimum force required. Understanding these relationships is key to solving the problem accurately.
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Homework Statement


is part C lack of information ? can someone give me a hint how to start ?

for part a ,
-(1.35-0.95)(200cos60)-2(200sin60)=-386N(anticlockwise)

for part b , -386.4/2 = 193N (in upward direction)

Homework Equations

The Attempt at a Solution

 

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There is enough information.
Suppose you resolve the tension into a force parallel to AB and a force perpendicular to it. What can you say about how these components contribute to the moment about A?
 
haruspex said:
There is enough information.
Suppose you resolve the tension into a force parallel to AB and a force perpendicular to it. What can you say about how these components contribute to the moment about A?
If i resolve the tension into a force parallel to AB and a force perpendicular to it , then I need to take the perpendicular distance multiply the force to get the moment . But , the angle between the force and x , y -axis is unknown , how to determine the force ? how to know it's minimum ?
 
werson tan said:
If i resolve the tension into a force parallel to AB and a force perpendicular to it , then I need to take the perpendicular distance multiply the force to get the moment . But , the angle between the force and x , y -axis is unknown , how to determine the force ? how to know it's minimum ?
What is the moment about A of the component parallel to AB?
There is enough geometrical information to find the angle ABF.
 
haruspex said:
What is the moment about A of the component parallel to AB?
There is enough geometrical information to find the angle ABF.
why angle ABF is required ?
 
werson tan said:
why angle ABF is required ?
You are asked to find the smallest force that could be applied at B in order to produce the same momemt about A that the force along BF has. I suggested resolving the force along BF into a component parallel to AB and a force perpendicular to AB. In order to do that you need to find the angle which BF makes to AB.
 
haruspex said:
You are asked to find the smallest force that could be applied at B in order to produce the same momemt about A that the force along BF has. I suggested resolving the force along BF into a component parallel to AB and a force perpendicular to AB. In order to do that you need to find the angle which BF makes to AB.
well, i still don't knw what do you mean here, btw , i managed to get the angle of ABF = arc(tan 0.4/2) = 108.7
 
What direction must the minimum force at B have in relation to line AB?
 
insightful said:
What direction must the minimum force at B have in relation to line AB?
Sorry, I am nt sure
 
  • #10
werson tan said:
Sorry, I am nt sure
Exercise: Take a 1m bar and apply a 10N force to one end. In what direction will the force produce the maximum moment at the other end of the bar?
 
  • #11
werson tan said:
well, i still don't knw what do you mean here, btw , i managed to get the angle of ABF = arc(tan 0.4/2) = 108.7
Ok, so now you know how to resolve the applied force F into components parallel to AB (call this FAB) and perpendicular to AB (FP). What moment does FAB have about A?
 
  • #12
insightful said:
Exercise: Take a 1m bar and apply a 10N force to one end. In what direction will the force produce the maximum moment at the other end of the bar?
max moment will produce when the force is applied exactly 1m from the point .
 
  • #13
haruspex said:
Ok, so now you know how to resolve the applied force F into components parallel to AB (call this FAB) and perpendicular to AB (FP). What moment does FAB have about A?
this looks weird, this is the first time i resolve the force with the angle more than 90 degree
 

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  • #14
werson tan said:
this looks weird, this is the first time i resolve the force with the angle more than 90 degree
That's the right diagram, but you did not answer my question. What is the moment of the FAB component about A?
 
  • #15
werson tan said:
max moment will produce when the force is applied exactly 1m from the point .
That is correct, but I was asking about the direction of the force at that point. For example, in this drawing, which direction of the 10N force (A, B, or C) produces the maximum moment at point O?:
Moments.jpg
 
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  • #16
insightful said:
That is correct, but I was asking about the direction of the force at that point. For example, in this drawing, which direction of the 10N force (A, B, or C) produces the maximum moment at point O?:View attachment 90160
B
 
  • #17
werson tan said:
B
Correct! So, what angle to any moment arm would give the minimum force to produce a given moment?
 
  • #18
insightful said:
Correct! So, what angle to any moment arm would give the minimum force to produce a given moment?
90 degree is to produce max force , how to determine the min force ?
 
  • #19
werson tan said:
90 degree is to produce max force , how to determine the min force ?
No, 90 degrees produces the maximum moment for a given magnitude of force; which is the same as saying it requires the minimum magnitude of force for a given moment.
 
  • #20
haruspex said:
No, 90 degrees produces the maximum moment for a given magnitude of force; which is the same as saying it requires the minimum magnitude of force for a given moment.
ok , i understand it . but , how to solve the question ?
 
  • #21
werson tan said:
ok , i understand it . but , how to solve the question ?
Use the diagram you posted at #13.
What I was trying to get you to see was that the component of F parallel to AB has no moment about A, so you might as well set it to zero. That leaves only the component perpendicular to AB. You know what moment this needs to have about A, and you know the distance AB, so what is the magnitude of this force?
 
  • #22
haruspex said:
Use the diagram you posted at #13.
What I was trying to get you to see was that the component of F parallel to AB has no moment about A, so you might as well set it to zero. That leaves only the component perpendicular to AB. You know what moment this needs to have about A, and you know the distance AB, so what is the magnitude of this force?
so , i have found out that the AB = 2.3m , this is due to sin11.3= 0.45/AB
the angle between the line AB and the horizontal line = 11.3 degree becoz i take 180 degree - 60 degree-108.7 degree , so i gt 11.3 degree

so i have found out the force = 386.4/ 2.3 = 168 N ... is it correct ?
 
  • #23
haruspex said:
Use the diagram you posted at #13.
What I was trying to get you to see was that the component of F parallel to AB has no moment about A, so you might as well set it to zero. That leaves only the component perpendicular to AB. You know what moment this needs to have about A, and you know the distance AB, so what is the magnitude of this force?
why the distance AB is required ?
 
  • #24
werson tan said:
so , i have found out that the AB = 2.3m , this is due to sin11.3= 0.45/AB
the angle between the line AB and the horizontal line = 11.3 degree becoz i take 180 degree - 60 degree-108.7 degree , so i gt 11.3 degree

so i have found out the force = 386.4/ 2.3 = 168 N ... is it correct ?
The 0.45 is wrong, leading to the wrong value for the minimum force.
 
  • #25
haruspex said:
The 0.45 is wrong, leading to the wrong value for the minimum force.
why the distance AB is required ? if i substituted 0.45 with 0.4 , then my ans would be correct ?
 
  • #26
werson tan said:
why the distance AB is required ? if i substituted 0.45 with 0.4 , then my ans would be correct ?
Yes, you would get the right answer.
The need to find distance AB should be evident from the calculation you have done. You knew what moment about A was to be provided by the force, and that the force was to be applied at point B. You found that the force would be at right angles to AB, so moment = magnitude x distance AB.
 
  • #27
haruspex said:
Yes, you would get the right answer.
The need to find distance AB should be evident from the calculation you have done. You knew what moment about A was to be provided by the force, and that the force was to be applied at point B. You found that the force would be at right angles to AB, so moment = magnitude x distance AB.
ok , i understand it nw. i used the method of resolving the forces to x and y components and then multiply them with the perpendicular distance. I have forgotten about the method moment = Fd
 

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