AAMC Physics Self Assessment #108 - Equilibrium

[moto_za]

A 0.5 kg uniform meter stick is suspended by a single string at the 30 cm mark. A 0.2 kg mass hangs at the 80 cm mark. What mass at the 10 cm mark will provide equilibrium?

Answer: 1kg

Looking at the AAMC solution they set it up as:

mg x 20cm = .2kg x 50 cm x g + 0.5 kg x 20 cm x g

I was wondering how you get the 20 cm on the right side of the equation? TIA!

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[volcomx]

A 0.5 kg uniform meter stick is suspended by a single string at the 30 cm mark. A 0.2 kg mass hangs at the 80 cm mark. What mass at the 10 cm mark will provide equilibrium?

Answer: 1kg

Looking at the AAMC solution they set it up as:

mg x 20cm = .2kg x 50 cm x g + 0.5 kg x 20 cm x g

I was wondering how you get the 20 cm on the right side of the equation? TIA!

mass of the stick can be treated as a point mass at the center of gravity (the middle), which is at the 50 cm mark, 20 cm from the string.

 

[moto_za]

mass of the stick can be treated as a point mass at the center of gravity (the middle), which is at the 50 cm mark, 20 cm from the string.

Thanks, how did you know the center is at 50?

 

[volcomx]

well, formally it would involve an integral involving density with respect to position. but the mcat would never test on that. you have to use common sense to realize that a 1 meter long stick would have the center of mass in the middle: 50cm.

 

[sazerac]

Thanks, how did you know the center is at 50?

Because the whole stick is one meter long?

 

[Temperature101]

Because the whole stick is one meter long?

How do you know that?

 

[volcomx]

How do you know that?

not sure if trolling or....

 

[Postictal Raiden]

How do you know that?

The question says "meter stick"

 

[Postictal Raiden]

the way I solve this is by setting all of the mass on the right side of the pivot point equal to those on the left side. The pivot point is at the 30cm mark, but let's call it zero for simplicity. Now we can the distance of masses accordingly.

Torques on the left = torques on the right

mass(x) 20cm from pivot = (center of mass)20cm from pivot + (0.2mass) 50cm from pivot

(20)(x) = (0.5)(20) + (0.2)(50)

20x = 20

x = 1

 

[Temperature101]

not sure if trolling or....

I was not trolling...I did not see or read meter stick..LOL...Then it is a simple problem.

 

[premed1001]

so whats the fastest way to approach this? i wasted alot of time trying to set this up

 

[Labkid]

so whats the fastest way to approach this? i wasted alot of time trying to set this up

I think you should draw out the problem. Usually I can solve most of the problems without having to write out an equation for them (and from what I've seen that's the case most times). They're trying to confuse you here and waste your time. You'll eventually get good enough to where you can visualize it in your head, but until then write the diagram out and things become a lot clearer.

Edit: Something like this

(Technically the masses don't equal it by themselves you have to remember to take distance from the center point into account. If these problems keep giving you trouble I suggest you study up a bit more on torque.)

 

[premed1001]

I think you should draw out the problem. Usually I can solve most of the problems without having to write out an equation for them (and from what I've seen that's the case most times). They're trying to confuse you here and waste your time. You'll eventually get good enough to where you can visualize it in your head, but until then write the diagram out and things become a lot clearer.

Edit: Something like this

(Technically the masses don't equal it by themselves you have to remember to take distance from the center point into account. If these problems keep giving you trouble I suggest you study up a bit more on torque.)

I will try that, I was trying to do it without a diagram before.
Thank you!