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Physics question...

Discussion in 'Pre-Medical - MD' started by AF1892, Aug 13, 2001.

  1. AF1892

    AF1892 Member
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    Hi, can someone please show me how to solve this question:

    An object with 15g mass is immersed in benzene and suffers an apparent loss of mass of 5g. What is the approximate specific gravity of the object? (Data: Specific gravity of benzene= 0.7)

    Thanks
     
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  3. Hopkins2010

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    Well I just glanced at this problem but what I'm thinking is:

    First draw a free body diagram of the object immersed in the benzene liquid. There are two forces acting on the object, a buoyant force and weight.

    Now, it states that the apparent weight of the object in immersion is 10*g, where g is the gravitational constant. According to the Law of Archimedes (or some crap like that) the buoyant force on an immersed object is equal to the weight of the displaced fluid. We need to find the density of the displaced benzene to resolve it into a force component, and we know that 0.7 = D g/cm^3 / 1 g/cm^3. Therefore, D = 0.7 g/cm^3.

    Since we arent given the volume of the object, I will just put another variable V into the equation to represent volume. Also, I'm assuming that the object is COMPLETELY immersed in the the benzene.

    Since those are the only two forces acting on the object (on a macro scale anyways), we have the following equation by Newton's Second Law:

    10g = 0.7Vg

    Simplifying, we have V = (10 g)/(.7 g/cm^3) = 14.3 cm^3.

    Now that we know what the total volume of the object is, we are ready for the final step.

    Originally we are told that the object's mass is 15g, so we can find the density of the object now that we know its volume.

    Specific gravity = ratio of density of object to density of water (1 g/cm^3).

    Thus, the specific gravity of the object = (15 g / 14.3 cm^3) / (1 g/cm^3) = 1.05

    Somebody please check my work in case I made a stupid mistake. There are shortcuts to this answer, I really blew this problem out of proportion by making it so detailed, but I thought that would be the best way to explain the problem, so you can use that line of thinking to solve more complex problems.
     
  4. Hopkins2010

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    On a side note, a mental note you might want to make to yourself on such problems is that any object that does not float must have a specific gravity greater than water. On the MCAT, it wouldnt surprise me if they put a couple of answer choices that are less than 1.0. Thus, you can eliminate a couple of answer choices just knowing basic info without using any equations.

    The only exceptions are in cases where the surface tension force is not superceded (i.e. feathers floating on water, water-striding insects, etc). These objects are special cases and the forces involved are different than for the problem above.
     
  5. Cobragirl

    Cobragirl Hoohaa helper ;)
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    Somebody PLEASE tell me I don't EVER have to do physics equations in medical school! PLEASE!!!!!

    (I can't believe I got through the MCAT...)

    PS- BaylorBoy, that was quite impressive!!
     
  6. moo

    moo 1K Member
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    Yes, I think you are correct baylor. If there's one thing I hate in physics (and there isn't many!) it's fluids. They just don't seem to teach that stuff to us.
     
  7. Hopkins2010

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    Cobragirl, you wont have to do any more physics equations ever again unless you want to get into biomedical imaging or diagnostic radiology perhaps.

    Trust me, simple fluid mechanics is a walk in the park compared to the calculus involved in MRI. I thought I had been exposed to alot of high level mathematics being in engineering, but I had seen nothing until I started MRI research.
     
  8. kreno

    kreno Candy Man
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    Correct me if i'm wrong, but shouldn't the equation be

    5g = .7Vg?

    I say 5 because i'm assuming you are saying that the buyant force equals the weight of the fluid displaced (like you said above). hence, 5g would equal the weight of the fluid displaced (i.e. .7Vg). Solving, your V would be.... 5/.7 = V. What is that, like 7 something? So, your specific gravity is 15/7something / density of water (which is one)... thus, specific gravity is 'bout 2?
     
  9. Legend

    Legend Super Senior Member
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    I think you are correct.

    5=0.7V
    V=7.1
    15g=p7.1g
    p=2.1
     
  10. Hopkins2010

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    Yeah, I see what you are saying.

    I understand why your method is correct, but I dont know why my method is wrong.

    I'll have to think about why you cant use a force diagram to solve this problem.
     
  11. Legend

    Legend Super Senior Member
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    baylor, I think your method is correct except that you seem to have a small numerical error.

    my short equations above are actually derived from your force diagram. There are only two forces in the system: Mg and Buoyant force (which is equivalent to 5g) acting in opposite direction.
     
  12. kreno

    kreno Candy Man
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    yep, baylor, you were right except for the one number. basically, remember... buyant force equals 5g... i think the question is worded weird, that's all
     
  13. Hopkins2010

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    Upon re-reading the question, I realized that I misread it and plugged the wrong number in.

    Just goes to show you, MAKE SURE YOU READ THE PROBLEM STATEMENT TWICE. :oops:

    Also, instead of using my method, its much faster to just remember the following principle: The buoyant force = weight of displaced fluid = apparent weight loss of the object.

    Thats a much more direct and quick way to get to the answer.
     
  14. MacGyver

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    Buoyancy problems are best solved by NOT using force diagrams.

    The reason for this is because force diagrams (drawn by layment) generally imply a static condition. Theres a whole course in engineering devoted to this topic called Statics.

    However, buoyancy problems are NOT statics problems because if the specific gravity is greater than 1.0, it will cause the object to sink. Therefore, a static analysis (i.e. static force diagrams) can't be used here.

    Now, if you want to use Dynamic force diagrams (moving objects) then you may do so. However, this tends to be a messy way to solve the problem.

    A sinking object has acceleration, so when drawing force diagrams you would have to account for this. Since the vast majority of buoyancy problems give you no information on acceleration, its better to just use the advice in the post above about the weight loss of the object being equal to the weight of the fluid displaced.

    A proper dynamic force diagram for this problem would have 2 forces, a 10g weight force acting downwards and a 0.7Vg buoyant force acting upwards.

    The reason its 10g acting downwards is because there is an "apparent mass loss" of 5 grams. This means that if you were to hook up a scale to the object immersed in the benzene, the scale would read 15g - 5g = 10g.

    Now, you cant ignore acceleration in this case, so using the dynamic force diagram you would get the following equation:

    10g - 0.7Vg = ma, or 10g - 0.7Vg = 15a

    Quickly we run into a problem here because to compute the acceleration of a rigid body in liquid is no easy task. You would have to use equations to first find the coefficient of kinetic friction in liquid and the effective cross-sectional area of the object exposed to the frictional force.

    You can do this, but it takes several equations and in general is not a very nice calculuation to do.

    If you go thru the methods, you get a specific gravity of 2.085 or so.
     
  15. groundhog

    groundhog 1K Member
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    My health watch nurse sent me a note. She said that I am overdue for a scope job. If I bite the bullet and go in, I sure hope the folks who man the equipment understand the general principles of mass, density, force, and acceleration.
     
  16. Bruin4Life

    Bruin4Life Senior Member
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    The people above are correct, the answer is approx 2.1. The one piece of advice that helped me when working through physics was: ORGANIZATION, and CONCEPTUAL UNDERSTANDING.

    I could've pulled out equations and started plugging and chugging, but that's the least efficient method for this type of problem.

    I answered the question above in less than one minute with a pen and a small piece of paper I ripped off an AMCAS printout that I had on my desk (**** those AMCAS applications have to come in handy somehow). Here's how I approached it:

    1.) What the hell is it asking? Ok it's asking for the specific gravity of an object. [Memory bank: Specific gravity of an object is simply its density divided by the density of water. You should know the density of water is 1g/ml] THE QUESTION NOW TURNS INTO: FIND THE DENSITY OF THE BLOCK.
    2.) What are the units of density? They're grams/volume. The grams of the block is given: 15 grams. THE QUESTION NOW TURNS INTO: FIND THE VOLUME OF THE BLOCK.
    3.)How do I solve for the volume? This is where you review bouyancy! I'll stop at the trickest step. The only way to learn is to sweat through problems.

    In high school I never took any science class aside from biology, and in college I majored in Psychology. I still pulled of a 10 on PS. I know that's not high, but it's good enough. Anyway, on my MCAT form, I didn't use more than a couple equations, it was conceptually inclined. So to the OP, refine your understanding of Physics/Chem concepts, it gets easier that way, as opposed to plugging into equations blindly.
     
  17. BCgirl

    BCgirl Member
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    When I wrote AAMC IV on Sunday, I answered this one by thinking that the liquid supported 1/3 of the weight of the block. If the liquid supported 1/3 of the weight, then the block's density (and specific gravity) would be 3 times that of the liquid. So, the specific gravity is 2.1. Can you think of it that way, or was I just lucky?
     
  18. kreno

    kreno Candy Man
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    why yes you can - in fact that's the easiest way to figure it out.
     
  19. Amy B

    Amy B I miss my son so much
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    I HATE PHYSICS!!!!!!!!!!!!!!!!!!!!
    !!!!!!!!!!!!!!!!!!!!!!!!!!!
    !!!!!!!!!!!!!!!!!!!!!!!!!!!
    !!!!!!!!!!!!!!!!!!!!!!!!!!!
    !!!!!!!!!!!!!!!!!!!!!!!!!!!
    :mad: :mad: :mad: :mad:
     

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