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Physics 19 Online
OpenStudy (anonymous):

The force F1, 10.0N acts at 10.0cm. What is the magnitude of the torque due to F1 about an axis through Point A perpendicular to the page? Is it clockwise,or is it counterclockwise? ( Point A is at 0 on a meter stick. There is also a point B at 50cm and F2 is 70cm with a force of 15.0N)

OpenStudy (anonymous):

what's the angle of F1 to the meter stick?

OpenStudy (anonymous):

no angle is given

OpenStudy (anonymous):

|dw:1349156959118:dw|

OpenStudy (anonymous):

so does it look like that?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

that's 90 degrees torque = F*r*sin(90)

OpenStudy (anonymous):

=10N*.1m*1 = 1Nm

OpenStudy (anonymous):

direction of torque would be into the screen (clockwise rotation)

OpenStudy (anonymous):

questions?

OpenStudy (anonymous):

what is the r? Is that the 10cm?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Hi @Algebraic! your ava's still gloomy as usual.

OpenStudy (anonymous):

sad, not gloomy.

OpenStudy (anonymous):

b/c math is hard :(

OpenStudy (thivitaa):

torque = rF = 0.1(10) =1 Nm

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