How did my professor know to convert 2sin(45) to = F/ (sq. rt.(2) ? I know that

Question

How did my professor know to convert 2sin(45) to = F/ (sq. rt.(2) ?

I know that 2sin(45) and the square root of 2 are equivalents, but how did she know to automatically put square root of two there?

Also, when she first starts to solve the problem, the puts the counterclockwise component of the force to equal LTsin45...why can't it just be LT, so that for the equation Torque = r F, that r can be L and F is T?

Thanks!

oNof 45 puot pt 2 2 owius of rottion (a) Let L= the rod length. The blue arrow represents the bolt force, F u ut 2sin 45° L77 bolt pts into Q2 SoY,Component is negrtNe (b) The vertical component of force at the bolt is equal, and opposite to, the V2cos49.ETgtot at the bolt can be found from FTu-bot cos 45° vertical component of the wire tension: The horizontal component of force at the bolt can be found from0 center ofrotation at the top of the rod: 2r-z-and 0 with the center of rotation at the top of the rod:and -

Explanation / Answer

2sin(45) F to = F/ (sq. rt.(2) ....this is wrong

2sin(45) F to = F*(sq. rt.(2)

as sin45 = 0.707 OR equavalently 1/(sqrt(2)

=> 2sin(45) F = 2* 1/(sqrt (2)) F

now 2 = (sqrt(2)*(sqrt(2)

=> 2sin(45) F = 2* 1/(sq. rt.(2) F = (sq. rt.(2)*(sq. rt.(2) * 1/(sq. rt.(2) F = (sqrt (2)) F

b)

torque = rxF ( it cross product between r and F) if u don't know what is that then only understand it is not the normal product = r=L, F= T => LT ( it is wrong)

torque = rxF = rFsin(theta) ( where theta is angle blw force vector and line joing to the point where force is acting which is 45 in this case)

=> Torque = LTsin(45)

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