EK 1001 physics 302 torque

[GRod18]

---

Members do not see this ad.

A one meter board with a mass of 12 Kg hangs by a rope as shown. A block is hung from the left end of the board. If the board is to be balanced level to the ground, what must be the mass of the block?

Picture is a long board with mass hanging at its left end, which is 0.3 meters away from the rope that is holding up the long board (rope is labeled T)

EK solved it as Torque counterclockwise = torque clockwise.
They picked the rope as the point of rotation and set 0.3 mg = 0.2 (12)g. It also says that the weight of the board acts at its center of gravity 0.5 m from the left end.

What I don't get is what this last statement means, also why did they use 0.2 (12) g and not 0.7 since thats on the other side of the rope??

 

[fizzgig]

masshang_down--0.3m--T_up--0.2m--ctrofmassofboard_down--0.5m--endofboard

so the board looks like that.

pick rotation around the tension point. now you can ignore that force. the block is 0.3m away, and the center of mass of the board is 0.2m away from this point of rotation (since the center of mass is at 0.5m). you don't use 0.7 because that assumes that the whole 12kg of the board is concentrated at the far end of it. it's a distributed weight, and if you want to use one vector to represent the weight of an object you must put it at the center of mass.

about your tension point, the block will tend to rotate the board one way, and the center of mass of the board will tend to rotate it the other. these are the only forces (besides tension, which we eliminated) acting on the board. set them equal.

you don't have to pick the tension point as the rotation point, it's just the fastest. i'm not the fastest study so i picked the far end of the board. and i had
Fblock(1) + Fboard(0.5) - Ftensn(0.7) = 0 and i knew Ftensn was the sum of the block and board weights. it's fine.

you could pick any arbitrary point to sum the torques.

 

[GRod18]

masshang_down--0.3m--T_up--0.2m--ctrofmassofboard_down--0.5m--endofboard

so the board looks like that.

pick rotation around the tension point. now you can ignore that force. the block is 0.3m away, and the center of mass of the board is 0.2m away from this point of rotation (since the center of mass is at 0.5m). you don't use 0.7 because that assumes that the whole 12kg of the board is concentrated at the far end of it. it's a distributed weight, and if you want to use one vector to represent the weight of an object you must put it at the center of mass.

about your tension point, the block will tend to rotate the board one way, and the center of mass of the board will tend to rotate it the other. these are the only forces (besides tension, which we eliminated) acting on the board. set them equal.

you don't have to pick the tension point as the rotation point, it's just the fastest. i'm not the fastest study so i picked the far end of the board. and i had
Fblock(1) + Fboard(0.5) - Ftensn(0.7) = 0 and i knew Ftensn was the sum of the block and board weights. it's fine.

you could pick any arbitrary point to sum the torques.

oh ic makes a lot of sense, I don't think the EK text explained this that well, and I don't remember doing problems with massed boards from TPR lol. thank you so much again!

" you don't use 0.7 because that assumes that the whole 12kg of the board is concentrated at the far end of it. it's a distributed weight, and if you want to use one vector to represent the weight of an object you must put it at the center of mass" this statement is on point wow! your going to own the physics section!