- #1

- 5

- 0

TLsinO - mg(1/2LcosO) = 0

Your help would be much appreciated.

- Thread starter fldk31
- Start date

- #1

- 5

- 0

TLsinO - mg(1/2LcosO) = 0

Your help would be much appreciated.

- #2

- 941

- 394

Do you know how to calculate a torque?

- #3

- 5

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t=fd and t-wa ?

- #4

- 941

- 394

What does d represent there?

- #5

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distance....

- #6

- 941

- 394

- #7

- 5

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I am so confused what these two parts mean:

1) TLsinO

2) mg(1/2LcosO)

- #8

- 35,136

- 6,271

If you are familiar with torque then you should know this: if a force F acts through a point P, the torque the force has about a point Q is F x distance PQ x sin(angle between PQ and the direction of the force).

I am so confused what these two parts mean:

1) TLsinO

2) mg(1/2LcosO)

Can you relate that to the diagram and see how the expressions you quote correspond to the torque exerted by T about the hinge, and the torque the weight of the arm exerts about the hinge?

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