friction and rolling resistance, and work done queriesby sgstudent Tags: rolling, work energy theorem 

#19
Apr112, 06:15 AM

P: 637

But then again, when the driving wheel turns looking at it from the right hand side, it turns clockwise, but my force is the opposite direction... thanks for the help tinytim! btw, i also have a dynamics question to ask hope you can help out there too! Thanks 



#20
Apr712, 10:28 AM

P: 637





#21
Apr712, 04:50 PM

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i said that in this case the work done against gravity is the same … so in the general case the (difference in) gravitational energy would have to be included 



#22
Apr712, 11:04 PM

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#23
Apr812, 05:50 AM

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hi sgstudent!
but you can either include it as a force (on the work done side), or you can include its potential energy (on the energy side), but not both!! as jtbell and Doc Al said seven(!) years ago … if the force F is not constant (or if the displacement is not straight), then instead of F.x we must use ∫ F.dx 



#24
Apr912, 01:31 AM

P: 637

Oh so if an object moves up an incline with constant speed, is the network done 0 or is it the gain in GPE? Thanks for the help!




#25
Apr912, 03:14 AM

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exam questions are usually about the work done by a particular force
they usually aren't interested in the net work done (since that will usually be zero, except in the case of friction, when net work done will equal loss of mechanical energy, ie gain in thermal energy) 



#26
Apr912, 04:45 AM

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#27
Apr1012, 05:36 AM

P: 637

I think it should work right? Since when I do work done by applied force and minus away all the negative works like wd against friction and gravity so the gravity part is cleared away by minus it already. Is this right? Thanks for the help!




#29
Apr1012, 08:30 AM

P: 637

thanks tiny tim! I think it makes more sense to include the work done against gravity rather than to leave it to a gain in GPE. Thanks again for the help!




#30
Apr1312, 06:13 AM

P: 637

hi tiny tim, i was thinking about the problem with the wheels again so for the nondriving wheels shouldn't the friction be forwards also? Since by torque they are both having the same direction ( as in turning direction)
also, when principle of moments occurs where there is no net moment where clockwise moment=anticlockwise moment can there be rotation? Like even constant speed rotation or will there be always no net moment? thanks for the help! 



#31
Apr1312, 06:32 AM

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the direction of friction is decided by which one is too big for the otherfor the driving wheels, there was angular acceleration from the engine for the nondriving wheels, there is no angular acceleration from the engine 



#32
Apr1312, 07:00 AM

P: 637

Hi tiny Tim! thanks for the help all this while so for the driving wheel is positive work being done? Doc Al said no here: http://www.physicsforums.com/showthread.php?t=92895 but in your previous post it was otherwise..
Also I came across a weird moments question. They asked me to find the moment about a point to lift a trap door. I'm not sure if I can use principles of moment here cos lifting it requires a net moment but then again constant speed moment can also occur. So I'm sure how does this work.. Thanks for the help tiny Tim you rock! 



#33
Apr1312, 07:50 AM

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(eg static friction on a rolling driving wheel is what accelerates the car) the minimum force necessary is the force that gives zero acceleration (anything less doesn't open the trap door, anything more isn't the minimum) (the net moment is of course the applied moment plus the moment of the weight) 



#34
Apr1312, 09:03 AM

P: 637

Hi Tim! it makes more sense this way. So when a car travels at constant speed is the forward friction equal to the backwards friction (non driving wheels) then when the forward wheel's friction is equal to the turning of the wheel such that the forward friction is equal to the turning force at the point of contact? Will the backwards resistance change accordingly since it is still static friction at the backwheels?
Also how do I tell when a moment will travel at constant speed or stay motionless when a equilibrium occurs similarly how do i tell when net force is 0 whether it is constant or 0 speed. Thanks for the help 



#35
Apr1312, 09:48 AM

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at constant speed, there is no turning force from the engine (i'm assuming we're ignoring air resistance etc) 



#36
Apr1312, 12:04 PM

P: 637

Like when I pull that trapdoor the net moment is 0 but there is constant speed while when a ruler is place on a pivot and two objects resting on both sides and moment is 0 it there is no movement of the plank even though in both cases net moment is 0 so how do I tell if an plane will move with constant speed or 0 speed at all. Thanks for the help!



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