Maximum torque

  • Thread starter amy098yay
  • Start date
  • #1
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Homework Statement


2. Homework Equations [/B]
I was given this info
r=1.5 m
d=3 m
F= 50
angle= 110 degrees


The Attempt at a Solution


T=Fsin
T= (1.5)(50)sin(110)
T=70 N

I attempted the solution^ to get (maximum) torque.
 

Answers and Replies

  • #2
23
0
i am unsure if my answer is correct or not*
 
  • #3
SteamKing
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You need to specify units for all the given quantities.

The units of torque are not newtons. What are they?
 
  • #4
23
0

Homework Statement


2. Homework Equations [/B]
I was given this info
r=1.5 m
d=3 m
F= 50 N
angle= 110 degrees


The Attempt at a Solution


T=Fsin
T= (1.5)(50)sin(110)
T=70 J (joules)

I attempted the solution^ to get (maximum) torque.
 
  • #5
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
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Although mathematically torque and work have the same units, the convention is to use units of F*L when talking about torque, and joules when talking about work.
 
  • #6
23
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thank you for notifying me about the units being used in this problem, so how would i go about getting MAXIMUM TORQUE?
 
  • #7
SteamKing
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thank you for notifying me about the units being used in this problem, so how would i go about getting MAXIMUM TORQUE?
Change the angle of application of the force.
 
  • #8
23
0
would it be 180-110=10 degrees
so
T=Fsin
T= (1.5)(50)sin(10)
T=13.0 J (joules) ???
 
  • #9
SteamKing
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would it be 180-110=10 degrees
Are you sure about this calculation?

so
T=Fsin
T= (1.5)(50)sin(10)
T=13.0 J (joules) ???
Think about it like this For what angle is the sine a maximum?
 
  • #10
23
0
90?

T=Fsin
T= (1.5)(50)sin(90)
T=75 (joules)
 
  • #12
gneill
Mentor
20,816
2,792
CWaters is correct, this thread duplicates a question in another thread. This thread is closed.
 

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