Torque and Fulcrum Problems :(

[premedicine555]

I'm doing passage II, question #8 of the Momentum/Torque section in TBR.. I;'m really confused about the rod balacing problems (there was a similar one in EK);

okay, so the first thing we have to do is "choose a pivot point"--which I did (I always choose the fulcrum as the point). And I normally measure the forces from the fulcrum... But whenever the rod's mass is taken into consideration (because there's no other force applied at the other end of the rod), things get confusing... In #8, why does the center of mass of the rod as a whole get taken into consideration..? Shouldn't setting the fulcrum as the pivot point be the only reference point we base the other masses on? *head desk*

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[Erythropoietin]

I'm doing passage II, question #8 of the Momentum/Torque section in TBR.. I;'m really confused about the rod balacing problems (there was a similar one in EK);

okay, so the first thing we have to do is "choose a pivot point"--which I did (I always choose the fulcrum as the point). And I normally measure the forces from the fulcrum... But whenever the rod's mass is taken into consideration (because there's no other force applied at the other end of the rod), things get confusing... In #8, why does the center of mass of the rod as a whole get taken into consideration..? Shouldn't setting the fulcrum as the pivot point be the only reference point we base the other masses on? *head desk*

I have the 09 version of the TBR book and couldn't find the #8 on the book but anyway, from what I understand... usually the rods are considered massless. When something is on a balance, you check for both translational and rotational equilibrium. In terms of rotational equilibrium, the mass of the rod wouldn't really matter much, because you are only dealing with r*F (which is torque) from both sides of the fulcrum. But when you are dealing with translational motion, instead of considering the left and right sides of the fulcrum, you now have to deal with up and down forces acting on the rod. If you have 10 N on both sides of a massless rod, your total applied downward force will be 20 N. But if you have a rod that has a mass of like 30 N, then now you have a total force of 50 N downwards. Obviously you end up with two different numbers when mass of the rod is taken into consideration. At the point where fulcrum touches the rod, it applies a normal force equal but opposite in direction of the weight of the boxes (Newton's 3rd law). So if you have a total mass of 10N + 10N + 30N = 50N, the normal force upward will be 50N (instead of 20N on a massless rod).

Again, I don't know which question you are referring to but I hope this helped somewhat ..

 

[deleted393595]

I'm doing passage II, question #8 of the Momentum/Torque section in TBR.. I;'m really confused about the rod balacing problems (there was a similar one in EK);

okay, so the first thing we have to do is "choose a pivot point"--which I did (I always choose the fulcrum as the point). And I normally measure the forces from the fulcrum... But whenever the rod's mass is taken into consideration (because there's no other force applied at the other end of the rod), things get confusing... In #8, why does the center of mass of the rod as a whole get taken into consideration..? Shouldn't setting the fulcrum as the pivot point be the only reference point we base the other masses on? *head desk*

I don't have the book, but could moment of inertia explain it?

 

[premedicine555]

Okay, the question states: The rod has a mass of 0.5kg. The counterweight has a mass of .01kg. With nothing on the pan, the rod is balanced at x=1cm. The counterweight (to the right of fulcrum) can be moved to balance out the torques and in doing so determine the weight of the idem added to the pan. What is the mass of the pan?

I know it's Torque(clockwise)=Torque(CC)... but TBR said that they also inputted the rod's torque, though they measured it from its center of mass.

 

[premedicine555]

The equation TBR gives is :10cm*m*g=1cm*.01kg(g) + 10cm*0.5kg*(g). then we solve for "m"; however, I don't understand the last part of the equation 10cm*0.5kg*(g).. why is it 10cm?? isn't the rest of the rod 30cm away from the fulcrum? Why is the "10cm' used?

 

[Pisiform]

The equation TBR gives is :10cm*m*g=1cm*.01kg(g) + 10cm*0.5kg*(g). then we solve for "m"; however, I don't understand the last part of the equation 10cm*0.5kg*(g).. why is it 10cm?? isn't the rest of the rod 30cm away from the fulcrum? Why is the "10cm' used?

Since the rod is uniform, the center of mass (or the mass of the entire rod) which is 0.5kg passes through the midway - means middle point of the entire rod's length i.e it will pass at 20cm. The pivot or fulcrum is 10cm from the left. The mass of the rod is 20cm from the left (or right ..midway). So the distance between pivot and the mass of the rod becomes 20-10 = 10cm.

Note that we are only interested in distance between the force (here, weight of the rod) and the pivot (which is 10cm from the left)

 

[premedicine555]

I guess I was mainly confused that the weight of the rod can be counted as another force.

So when it says that the "rod is balanced" that means that the rod's center of mass is ALWAYS in the middle of the rod (so we can divide 40/2)?

 

[Erythropoietin]

I guess I was mainly confused that the weight of the rod can be counted as another force.

So when it says that the "rod is balanced" that means that the rod's center of mass is ALWAYS in the middle of the rod (so we can divide 40/2)?

If you have unequal weights on both ends of the rod, the center of mass will NEVER be in the middle of the rod. It will only be "always" in the middle as long as there are equal masses on both sides of the rod.

 

[Pisiform]

I guess I was mainly confused that the weight of the rod can be counted as another force.

So when it says that the "rod is balanced" that means that the rod's center of mass is ALWAYS in the middle of the rod (so we can divide 40/2)?

Balanced does not necessarily means that the center of mass is in the middle. You can also balance a non uniform rod. But if the rod is uniform, then the center of mass is always in the middle. Rods in questions are almost always uniform unless otherwise stated