
#1
Aug511, 05:00 PM

P: 317

1. The problem statement, all variables and given/known data
A 1.0 kg rock is suspended by a massless string from one end of a 1.0 m long measuring stick. What is the mass of the measuring stick if it is balanced by a support at the 0.25 m mark? 2. Relevant equations Torque=(distance perpendicular to force from the center of mass)(mass of force)(acceleration of force) Torquesum=0 in static equilibrium 3. The attempt at a solution Torquesum=0= +(counterclockwise)(.25meters)(9.81m/s^2)(1kg) + (counterclockwise)(?.5meters?)(.25total mass Stick)(9.81m/s^2)  (clockwise)(?.5meters?)(.75TMS)(9.81m/s^2) solve for TMS and 2.4525=2.4525TMS; TMS=1kg So, why is ?.5meters? considered to be how far away the meter stick is from the center of mass? Originally, the individual centers of masses I tried to put as .25meters for the left part of the stick and .75meters for the right part of the stick. How did I make a mistake? 



#2
Aug511, 05:46 PM

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The centre of mass of the measuring stick is at the middle. The weight of the stick acts at that point. I do not understand why do you cut the stick into two.
The torque can be calculated with respect to any point, but it easiest to calculate it with respect to the support. ehild 



#3
Aug511, 05:49 PM

P: 317

oh thank you. then the center of mass with respect to the stick itself is .5meters which is why you use that quantity to calculate torque right?




#4
Aug511, 05:56 PM

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Torque (Need help with Center of Mass)
Yes, the centre of mass of the stick is 0.5 m from one end. If you calculate the torque with respect to the support, you have to use 0.25 m.
ehild 



#5
Aug511, 06:15 PM

P: 317

thats only for the rock though right?




#6
Aug511, 08:53 PM

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What do you mean?
There are three forces: the tension of the string the rock is connected to, but it is equal to the force of gravity on the rock; the weight of stick at the CM of the stick; and the upward force of the support. Both the resultant of these three forces and the resultant of their torques is zero. You can calculate the torque with respect to any point. ehild 



#7
Aug511, 09:11 PM

P: 317

so the torque of the rock is (9.81)(.25meters from the pivot point)(1kg)
the torque of the left end of the stick is (9.81)(.25massofstick)(.5meters from center of mass) and the torque of the right end of the stick is +(9.81)(.75mass of stick)(.5meters from center of mass) and adding these up gets you 0 since its static equilibrium how else would you go about solving this torque? 



#8
Aug511, 09:23 PM

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If you cut the stick to pieces, the weight of each piece acts at the centre of that piece. How do you calculate the torque of a force? ehild 



#9
Aug511, 09:41 PM

P: 317

Torque=(distance perpendicular to force from the center of mass)(mass of force)(acceleration of force)




#10
Aug511, 09:50 PM

P: 317

I'm basically trying to ask why you'd use .5meters (distance from the center of mass) instead of using however amount of meters away from the pivot point. I understand now. Well, basically, since the pivot point is anything in relation to forces that are trying to circulate in such away that the fulcrum is the center for the rock and the center of mass for the stick is the middle of the stick itself.




#11
Aug511, 09:52 PM

P: 317

the only unanswered question now is how else would you go about solving this?




#12
Aug511, 09:58 PM

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Supposing a massless rod can move horizontally around a vertical axis, and a force F acts perpendicularly on the rod at d distance from the axis, what is the magnitude of the torque? ehild 



#13
Aug511, 10:00 PM

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What is the force exerted by gravity on the stick? Where does it act? How far is that "point of attack" from the pivot? What is the torque of the weight of the stick with respect to the pivot? ehild 



#14
Aug511, 10:06 PM

P: 317

mass of force meaning the mass that is being affected by a force which in this case is only gravity and acceleration of force which again is gravity.
Fdsin180=torque trust me. If you dont use .5meters, the answer is wrong. You should try and work out the problem to see where you get. I know I know...its not your problem, but you have to use .5meters. Idk how else it can be done. 



#15
Aug511, 10:07 PM

P: 317

i meant Fdsin90=torque duh




#16
Aug511, 10:14 PM

P: 317

+(9.81)(.25meters from the pivot point)(1kg)
+(9.81)(.25massofstick)(.5meters from center of mass) (9.81)(.75mass of stick)(.5meters from center of mass) 2.4525t+1.22625tmos3.67875tmos=0 mos=1kg how else would you solve this. this is in fact that correct answer. I thought it was something else at first, but I was wrong. This is the correct answer. 



#17
Aug511, 10:27 PM

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Well, the torque from the rock is 1 kg * g* 0.25 m counterclockwise,
The weight of the stick acts at the CM, middle of the stick, at 0.25 m distance from the pivot. Its torgue is clockwise, m *g *0.25. mg*0.25=g*0.25, m=1 kg. The equation you wrote up for the torque is wrong, but accidentally has lead the same result for m. ehild 


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