Why is torque Fd for rolling without slipping?

[Hernaner28]

Hi. I'm taking a look at this problem:

And my doubt is with step 2 when he calculates the torque. He just says it is Fd. But...why? F only has a component which effectivly does torque.

Thanks

 

[Infinitum]

Hi Hernaner28!

And my doubt is with step 2 when he calculates the torque. He just says it is Fd. But...why? F only has a component which effectivly does torque.

Thanks

F as a whole is the component which creates effective torque. Fn doesn't create torque as its line of action passes through the ball's center, and the same with mg. But F has a line of action that is at a distance 'd' away from the center and perpendicular to it. So...

 

[Hernaner28]

I don't understand. F has a tangential component and that's the one which creates torque, the other component of F doesn't create torque because it passes through the center. Maybe it's a geometric problem I cannot solve...

Thanks!

 

[Infinitum]

I don't understand. F has a tangential component and that's the one which creates torque, the other component of F doesn't create torque because it passes through the center. Maybe it's a geometric problem I cannot solve...

Thanks!

Yes. But both the torques, that is the torque taken with the tangential component, or just taken the way it is, give you the same result! That's because for the tangential component, the perpendicular distance from the center is R, but in the case of the force given, it is d. Think about it...

 

[Hernaner28]

Ahhmmm I see... I didn't know you could take that easy way.. Thanks!