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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1Force times distance ..... right?

lizki
 one year ago
Best ResponseYou've already chosen the best response.0yes ,but i find them hard when we also have cos37 and sin37!

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1thats part of vectors then, and not really torque ... im assuming is there a more specific problem that we can relate these to?

lizki
 one year ago
Best ResponseYou've already chosen the best response.0This example is solved in my book,but i cant understand the procedure! The force applied to a 0.2m long spanner is 15N.Find the torque produced by this force.(take cos37=0.8 and sin37=0.6)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1then tag me in your post so I can have a look see later please

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we want to find the part of the force that is perpendicular to the spanner

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dw:1361799368979:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the amount of force applied at a right angle to the spanner is therefore: 15 sin(37) and the distance is already given as : .2m you said?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1so your question is not about torque ... but rather how to find the required adjustment of the force.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1which is just your right triangle material :)

lizki
 one year ago
Best ResponseYou've already chosen the best response.0But the topic in the book is 'TORQUE' but anw,thenkyou :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1youre welcome; and yes, torque is the new material that they are adding to the old material from previous sections :) good luck

Iamgmg90
 one year ago
Best ResponseYou've already chosen the best response.0\[Torque = \overrightarrow{F} \times \overrightarrow{r} \times Sin \Theta \]
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