TBR Physics Chapter 4

[darkpassengerMD]

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How do you solve question 8? I'm not understanding the concept of it. I'm weak on torque and having difficulty with this chapter. 🙁 Help!

 

[ayocaptain628]

so the whole idea is that the counterweight can be moved so that the torque on the left equals the torque on the right and the rod is balanced.

so in order to find the mass of the pan when nothing is on it, you have to find the net torque on each side.
the "sides" are determined by the fulcrum, b/c need the distance of the lever arm

so the net torque on the left side is just the pan

(m)*g*(10cm) b/c it's 10cm away from the fulcrum
where m is the variable for the mass of the pan

the net torque on the right side is the sum of two torques, the counterweight's and the rod's:

(.01 kg)*g*(1cm) <- for the counterweight {1cm is used b/c thats whats provided in passage as the distance the counterweight had to be moved to balance the torques with NO mass on the pan}
and also something a little more abstract
the rod, itself, contributes torque b/c it has mass to it
so you have to find the center of mass relative to the fulcrum
since the rod is 40 cm, then the center of mass is 40/2 = 20 cm b/c it will be right in the middle
so since the center of mass is 20 cm from the left or from the right,
its distance from the fulcrum is 10cm .:.
the torque contributed by the rod is
(.5 kg)*g*(10cm) <- for the rod

these torques must be equal to each other to balance out
so m*g*(10cm) = (.01kg)*g*(1cm) + (.5kg)*g*(10cm)
m = ((.01kg)*g(1cm) + (.5kg)*g*(10cm))/(g*10cm)
m = .5kg once you do the math

 

[darkpassengerMD]

so the whole idea is that the counterweight can be moved so that the torque on the left equals the torque on the right and the rod is balanced.

so in order to find the mass of the pan when nothing is on it, you have to find the net torque on each side.
the "sides" are determined by the fulcrum, b/c need the distance of the lever arm

so the net torque on the left side is just the pan

(m)*g*(10cm) b/c it's 10cm away from the fulcrum
where m is the variable for the mass of the pan

the net torque on the right side is the sum of two torques, the counterweight's and the rod's:

(.01 kg)*g*(1cm) <- for the counterweight {1cm is used b/c thats whats provided in passage as the distance the counterweight had to be moved to balance the torques with NO mass on the pan}
and also something a little more abstract
the rod, itself, contributes torque b/c it has mass to it
so you have to find the center of mass relative to the fulcrum
since the rod is 40 cm, then the center of mass is 40/2 = 20 cm b/c it will be right in the middle
so since the center of mass is 20 cm from the left or from the right,
its distance from the fulcrum is 10cm .:.
the torque contributed by the rod is
(.5 kg)*g*(10cm) <- for the rod

these torques must be equal to each other to balance out
so m*g*(10cm) = (.01kg)*g*(1cm) + (.5kg)*g*(10cm)
m = ((.01kg)*g(1cm) + (.5kg)*g*(10cm))/(g*10cm)
m = .5kg once you do the math

Is there a specific formula for this? It looks like you are using potential energy? My concern is when I look at questions like these that are associated with balance and torque. What is the best way in tackling these questions? I have a hard time grasping it.

 

[ayocaptain628]

Is there a specific formula for this? It looks like you are using potential energy? My concern is when I look at questions like these that are associated with balance and torque. What is the best way in tackling these questions? I have a hard time grasping it.

ya a formula for torque in general is: T = rF where r is the distance from the fulcrum and R is the force
that is why it looks like potential energy
b/c the force in this case is weight for all of them
and r is the distance of the weight vector from the fulcrum
so we were substituting values in this problem for T = r*mg

when u see these type of problems you gotta think
"balance" aka no net torque aka left torque equals right torque

a parallel to this is like seeing a kinematics problem with wording like "constant velocity"
no acceleration so no net force meaning the sum of the forces must equal 0.

thats usually what torque problems are all about
"rod is balanced" .:. torques on left of fulcrum (point of rotation) must equal torques on right of fulcrum

 

[darkpassengerMD]

ya a formula for torque in general is: T = rF where r is the distance from the fulcrum and R is the force
that is why it looks like potential energy
b/c the force in this case is weight for all of them
and r is the distance of the weight vector from the fulcrum
so we were substituting values in this problem for T = r*mg

when u see these type of problems you gotta think
"balance" aka no net torque aka left torque equals right torque

a parallel to this is like seeing a kinematics problem with wording like "constant velocity"
no acceleration so no net force meaning the sum of the forces must equal 0.

thats usually what torque problems are all about
"rod is balanced" .:. torques on left of fulcrum (point of rotation) must equal torques on right of fulcrum

Thank you so much. That actually makes a lot of sense. I have just studying for the mcat. I signed up for January. I haven't taken Physics 2 yet so I don't know what the physics 2 material is like compare to physics 1 but in physics 1 i got an A and torque was the only chapter I was iffy about. I breezed thru chapter 1-3 and now as you can see I'm on four.

 

[ayocaptain628]

Thank you so much. That actually makes a lot of sense. I have just studying for the mcat. I signed up for January. I haven't taken Physics 2 yet so I don't know what the physics 2 material is like compare to physics 1 but in physics 1 i got an A and torque was the only chapter I was iffy about. I breezed thru chapter 1-3 and now as you can see I'm on four.

np! i actually have never taken my physics prereqs either so i had to self teach myself.
but i started with the physics 2 book of TBR b/c apparently those topics are all high yield (optics, e&m, fluids)
gl studying!

 

[darkpassengerMD]

np! i actually have never taken my physics prereqs either so i had to self teach myself.
but i started with the physics 2 book of TBR b/c apparently those topics are all high yield (optics, e&m, fluids)
gl studying!

Thanks bro it doesn't look bad. Just need to study it.
Good luck to you also.