Work & Momentum: T, F, r, sinθ

[Anas]

Hello,
#1 As my teacher explained "T=F•r•sin<" all clear so far.
When i attempted to apply the formula it crossed my mind that the unit of T is (N•m), if so how do we solve for distence 0< r < 1.
#2 Another question if the unit is {(kg•m)/(s^2)}•m why did he represent moment with Newtons?
Assuming that he follows a rule of equivalency from tork to Newtons, how is it the same value on two different units?

 

[sophiecentaur]

Hello,
#1 As my teacher explained "T=F•r•sin<" all clear so far.
When i attempted to apply the formula it crossed my mind that the unit of T is (N•m), if so how do we solve for distence 0< r < 1.
#2 Another question if the unit is {(kg•m)/(s^2)}•m why did he represent moment with Newtons?
Assuming that he follows a rule of equivalency from tork to Newtons, how is it the same value on two different units?

Hmm, you will need to tell us the context of this formula. As it stands, it could be to do with many different situations.

 

[jbriggs444]

Assuming that he follows a rule of equivalency from tork to Newtons, how is it the same value on two different units?

It is difficult to understand what you are asking. My guess is that you are wondering why the unit for torque has the same dimensions as the unit for energy, both having dimensions of mass times distance squared over time squared.

One answer is that both are the product of a force times a distance. Another is that one Newton-meter of work is just what you would expect when applying one Newton-meter of torque over one radian of rotation. Since the radian is dimensionless, the units for torque and work will have the same dimensions.