- #1

- 50

- 7

- Homework Statement
- Determine the moment of the force about point O.

- Relevant Equations
- M = F.d

There are components of 500N:

500cos(45)= 353.55

500sin(45)= 353.55

Radius is 3 then

M = (353.55*5.12) - (353.55*2.12) = 1060.65

is that correct?

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I had a momentary brain cramp. The torque is indeed counter-clockwise. Good catch!In summary, the conversation discusses the components of a 500N force, its horizontal and vertical components (500cos(45) and 500sin(45)), a radius of 3, and the calculation of the moment of the force about point O. The moment is determined by multiplying the vertical component by the horizontal moment arm and subtracting the product of the horizontal component and the horizontal moment arm. The final answer is 1060.65 Nm, with counter-clockwise rotation being the convention.f

- #1

- 50

- 7

- Homework Statement
- Determine the moment of the force about point O.

- Relevant Equations
- M = F.d

There are components of 500N:

500cos(45)= 353.55

500sin(45)= 353.55

Radius is 3 then

M = (353.55*5.12) - (353.55*2.12) = 1060.65

is that correct?

- #2

Homework Helper

Gold Member

- 2,615

- 1,114

It's always good practice to show (or state) your convention with the diagram, and show units in computation. Sig figs are probably too many as well. In this case you chose counter-clockwise as positive moment.Homework Statement:Determine the moment of the force about point O.

Relevant Equations:M = F.d

View attachment 324887

There are components of 500N:

500cos(45)= 353.55

500sin(45)= 353.55

Radius is 3 then

View attachment 324888

M = (353.55*5.12) - (353.55*2.12) = 1060.65

is that correct?

Computationally...the calculation is correct.

Last edited:

- #3

Science Advisor

Homework Helper

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- 4,543

The blue vector is ##{1\over 2}r\sqrt 2##. That times the 500 from ##F## is 1061 Nm

##\ ##

- #4

- 50

- 7

Thanks alot.It's always good practice to show (or state) your convention with the diagram, and show units in computation. Sig figs are probably too many as well. In this case you chose counter-clockwise as positive moment.

Computationally...the calculation is correct.

- #5

Science Advisor

Homework Helper

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So you are splitting the applied force into its vertical and horizontal components. You computed the torque from the vertical component by multiplying by the horizontal component of the moment arm for the point of application (M = (353.55*5.12) - (353.55*2.12) = 1060.65

You computed the torque from the horizontal component by multiplying against the horizontal moment arm to the same point of application (5.12,

That is a viable approach. Straight, by the book, crank and grind.

The approach that I took was different.

The torque from a given force is the same no matter where that force is applied, as long as the revised point of application is somewhere along the "line of action" of the original force.

The drawing makes it clear that the line of action passes through the point (3.00, 0). That simplifies the math. Now the vertical moment arm is zero and we need only consider the vertical force component of 353.55 and the horizontal moment arm of 3.00:$$3.00 * 353.55 = 1060.65$$By inspection, this is a

Last edited:

- #6

- 50

- 7

I think you wrong it's counter-clouckwise becuase in this case when we applied a force horizontal it's counter-clouckwise.The drawing makes it clear that the line of action passes through the point (3.00, 0). That simplifies the math. Now the vertical moment arm is zero and we need only consider the vertical force component of 353.55 and the horizontal moment arm of 3.00:By inspection, this is a clockwise torque.

- #7

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I agree.it's counter-clockwise

- #8

Science Advisor

Homework Helper

- 12,177

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You are right, of course.I think you wrong it's counter-clouckwise becuase in this case when we applied a force horizontal it's counter-clouckwise.

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