torque

[minutemen11]

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View attachment torque.pdf

I may be slightly stupid but I really dont understand why in the explanation for this problem, the pulley system has a clockwise torque... if its on the other side of the fulcrum wont its torque move counterclockwise?

i wasnt able to put this as an image because of sizing restrictions

 

[jaxasp]

Try balancing the forces mathematically:
Right side-> 2g*1m (meter is obv. arbitrary) + 4g*3m = 14
Left side-> 10g*2m=20
I'm assuming the question asked for what weight is needed to level the seesaw, so you would need to somehow subtract 6 from the left side, and the pulley would pull upward on the seesaw (subtracting weight). The pulley attaches 3m over, so its 3m*x = 6 -> x=2
Basically, because the torque is higher on the left side, and torque needs to be applied in the clockwise direction, which is how the pulley affects the seesaw.

 

[jaxasp]

just realized i never answered your actual question; if the pulley was instead a weight, then yes the force would be counterclockwise, but because the pulley is applying an upward force on the left side, it is a clockwise force. draw a force diagram, the pulley is applying a torque directly upward, which is clockwise

 

[minutemen11]

just realized i never answered your actual question; if the pulley was instead a weight, then yes the force would be counterclockwise, but because the pulley is applying an upward force on the left side, it is a clockwise force. draw a force diagram, the pulley is applying a torque directly upward, which is clockwise

haha yeah thanks, i knew the answer. it just didnt make sense to me qualitatively until i drew out the force diagram.. kinda tricky i guess