it's becuase the output is not postive 2 KW and 1 KW, It's -2KW and -1KW. So when you subtract the 2 KW from 3 KW, you get 1 KW as a result. I hope this helps
You should think how much power is used in every section. In section AB until B shaft, 3kW power is used. Since B shaft uses 2kW of power, only 1kW of power is left for section BC.
The “uh huh uh huh” part always kills me😂
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Wonderful solids lessons!! really taking the time to explain the problems along with a good sense of humor :))
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Thanks for putting out these lessons!! It has really helped me through my studies
man his good jokes are good
HE MAKES LEARNING FUN!
Fr though great video
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In cancelling out units when getting the Torque, what happened to the "rad" underneath?
Since the radian is a dimensionless unit (it's a ratio of two lengths), dividing by radians has no effect on the unit itself. So N*m/rads = N*m
Why is the value of P in section BC is 1000 Watt, while in section AB is 3000 Watt?
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yes i noticed this as well I think this is wrong.
it's becuase the output is not postive 2 KW and 1 KW, It's -2KW and -1KW. So when you subtract the 2 KW from 3 KW, you get 1 KW as a result.
I hope this helps
@@MMMMMMMMMMBALLShow can you tell if it is positive or negative?
@@470bubbles Watch again from 7:27 sec onwards
Oh dear lord Thank God for Jeff hansen my teacher makes no sense
why do you disregard the 2kW at Point B? how do you know when to do that?
You should think how much power is used in every section. In section AB until B shaft, 3kW power is used. Since B shaft uses 2kW of power, only 1kW of power is left for section BC.
Unit of Power should be Kgm²/s³
What does he mean by don't get "J" confused with "I"?
What the hell? This man?😂😂💔
Jeff Hanson the next coming of our lord and savior