How Accurate Are These Torque Calculations?

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The discussion focuses on torque calculations based on given parameters: radius (r=1.5 m), distance (d=3 m), force (F=50 N), and angle (110 degrees). The correct torque applied is calculated as T=70 Nm using the formula T=rFsin(angle). Maximum torque occurs at 90 degrees, yielding T=75 Nm. The conversation highlights the importance of clarity in calculations and the need for a diagram to visualize the problem accurately.

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amy098yay
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I was given this info
r=1.5 m
d=3 m
F= 50 N
angle= 110 degrees

1. What torque is applied?
T=Fsin
T= (1.5)(50)sin(110)
T=70 J

2. At what angle would enough force be applied to achieve maximum toque?
T=Fsin
T= (1.5)(50)sin(90)
T=75 (joules)

3. If someone continues to push with a force of 50 N, what maximum torque could be applied?
T=Fsin
T= (1.5)(50)sin(10)
T=13.0 J

Are these solutions to the questions correct, if not; what do i need to change to get the correct solution?
 
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What do distances r and d represent? Where is the angle measured? The problem statement is very vague. A diagram would help.
 
For 1
First, I would like to see a diagram, as I don't know what the angle is between.
Second, I would point out that torque is not measured in Joules. I can see how you got that, but the force and distance in a torque are perpendicular. The force is not moving through that distance.
For 2
You need to read the question again. You know the correct answer, but have said something else!
For 3
Where did 10o come from? Again, you know the correct answer, I think: when you sort 2, you'll answer 3.

Edit - I see now this is continued from another thread. I think you need to draw a diagram.
Maximum torque: torque is Force x distance from centre of rotation, so to get max torque you need to maximise these. If you can't change the force, when is the distance from centre max?
 
Last edited:
heres the diagram
 

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I was given this info
r=1.5 m
d=3 m
F= 50 N
angle= 110 degrees

1. What torque is applied?
T=Fsin
T= (1.5)(50)sin(110)
T=70 N m

2. At what angle would enough force be applied to achieve maximum toque?
T=Fsin
T= (1.5)(50)sin(10)
T=13.0 N m

3. If someone continues to push with a force of 50 N, what maximum torque could be applied?T=Fsin
T= (1.5)(50)sin(90)
T=75 N mAre these solutions to the questions correct, i fixed the things you told me to. if not; what do i need to change to get the correct solution?
 
Last edited:
1. What torque is applied?
T=Fsin
T= (1.5)(50)sin(110)
T=70 Nm

You get the right answer, but your working is not clear.
T <> F sin rather T = r F sin(x) (You did not mention r. sin(x) is just because I think sin has to be sin of something.)
T= (1.5)(50)sin(110) Now that's right, but you put in r which was not in your original formula. That is bad IMO.
T = 70 Nm which follows from the previous line and is correct.

Question 2 asks "What ANGLE <gives> maximum torque?" You don't answer with an angle.
Where did 10o come from? It is not in your diagram (and neither is r or D btw) and you have not calculated it, so you need to explain it.

Q3 is correct, but disagrees with Q2. You need to explain what you are doing /thinking. It will help you.
 
amy098yay said:
2. At what angle would enough force be applied to achieve maximum toque?

Is that the exact wording of the question? The bit about "enough force" is slightly strange. It sounds like we might be missing part of the problem statement.
 
Does 90 degrees give maximum torque for 2. ?
 
Yes.
 

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