- #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.

- Thread starter amy098yay
- Start date

- #1

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2. Homework Equations [/B]

I was given this info

r=1.5 m

d=3 m

F= 50

angle= 110 degrees

T=Fsin

T= (1.5)(50)sin(110)

T=70 N

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

- #2

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i am unsure if my answer is correct or not*

- #3

SteamKing

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The units of torque are not newtons. What are they?

- #4

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2. Homework Equations [/B]

I was given this info

r=1.5 m

d=3 m

F= 50 N

angle= 110 degrees

T=Fsin

T= (1.5)(50)sin(110)

T=70 J (joules)

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

- #5

SteamKing

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- #7

SteamKing

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Change the angle of application of the force.

- #8

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would it be 180-110=10 degrees

so

T=Fsin

T= (1.5)(50)sin(10)

T=13.0 J (joules) ???

so

T=Fsin

T= (1.5)(50)sin(10)

T=13.0 J (joules) ???

- #9

SteamKing

Staff Emeritus

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Are you sure about this calculation?would it be 180-110=10 degrees

Think about it like this For what angle is the sine a maximum?so

T=Fsin

T= (1.5)(50)sin(10)

T=13.0 J (joules) ???

- #10

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90?

T=Fsin

T= (1.5)(50)sin(90)

T=75 (joules)

T=Fsin

T= (1.5)(50)sin(90)

T=75 (joules)

- #11

CWatters

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https://www.physicsforums.com/threads/torque-problem-confused.800504/

- #12

gneill

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CWaters is correct, this thread duplicates a question in another thread. This thread is closed.

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