- #1
werson tan
- 183
- 1
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)
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 ?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?
What is the moment about A of the component parallel to AB?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 ?
why angle ABF is required ?haruspex said:What is the moment about A of the component parallel to AB?
There is enough geometrical information to find the angle ABF.
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.werson tan said:why angle ABF is required ?
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) = 108.7haruspex 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.
Sorry, I am nt sureinsightful said:What direction must the minimum force at B have in relation to line AB?
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?werson tan said:Sorry, I am nt sure
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?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
max moment will produce when the force is applied exactly 1m from the point .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?
this looks weird, this is the first time i resolve the force with the angle more than 90 degreeharuspex 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?
That's the right diagram, but you did not answer my question. What is the moment of the FAB component about A?werson tan said:this looks weird, this is the first time i resolve the force with the angle more than 90 degree
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?:werson tan said:max moment will produce when the force is applied exactly 1m from the point .
Binsightful 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
Correct! So, what angle to any moment arm would give the minimum force to produce a given moment?werson tan said:B
90 degree is to produce max force , how to determine the min force ?insightful said:Correct! So, what angle to any moment arm would give the minimum force to produce a given moment?
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.werson tan said:90 degree is to produce max force , how to determine the min force ?
ok , i understand it . but , how to solve the question ?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.
Use the diagram you posted at #13.werson tan said:ok , i understand it . but , how to solve the question ?
so , i have found out that the AB = 2.3m , this is due to sin11.3= 0.45/ABharuspex 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 ?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?
The 0.45 is wrong, leading to the wrong value for the minimum force.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 ?
why the distance AB is required ? if i substituted 0.45 with 0.4 , then my ans would be correct ?haruspex said:The 0.45 is wrong, leading to the wrong value for the minimum force.
Yes, you would get the right answer.werson tan said:why the distance AB is required ? if i substituted 0.45 with 0.4 , then my ans would be correct ?
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 = Fdharuspex 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.
"Min Force Needed: Part C Hint" is a phrase that refers to a specific part of a scientific experiment or research project that requires the minimum amount of force to achieve a desired result.
Determining the minimum force needed in a scientific experiment is important because it allows for the most efficient use of resources and can help save time and money. It also ensures that the experiment is conducted accurately and consistently.
The minimum force needed in an experiment can be determined through careful observation, measurement, and data analysis. This may involve conducting multiple trials and adjusting the force as needed to achieve the desired result.
Yes, the minimum force needed in an experiment can vary depending on the environment or conditions in which the experiment is conducted. Factors such as temperature, pressure, and humidity can all affect the minimum force needed to achieve a certain result.
Knowing the minimum force needed in a scientific experiment can have various applications, such as in the fields of engineering, physics, and materials science. It can also be useful in industries such as manufacturing, where efficiency and precision are crucial.