Solving Two Physics Problems: Weighing Rocks and Steel Barbells

AI Thread Summary
The discussion revolves around solving two physics problems involving buoyancy and density. The first problem requires determining the fraction of a rock specimen's volume that is solid, using Archimedes' Principle and the relationship between weight in air and water. The second problem involves calculating the force a bodybuilder must exert to lift a barbell underwater, also relying on buoyancy concepts. Participants express confusion about applying the relevant equations without knowing the mass directly. The conversation emphasizes the importance of understanding the principles of density, volume, and buoyant force in solving these problems.
SnowOwl18
Messages
71
Reaction score
0
Alright, these two problems are driving me nuts:

1. ---To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimen weighs twice as much in air as it does in water. The solid part of the specimen has a density of 5.10×103kg/m3. What fraction of the specimen's apparent volume is solid?----

I think that this involves Archimede's Principle because she floats the rock in water to find displacement... the equation is (magnitude of buoyant force)= (weight of displaced fluid) ...but I don't see how I can use that equation..

Also this problem:
----A bodybuilder is holding a 29.9kg steel barbell above her head. How much force would she have to exert if the barbell were lifted underwater?---

It doesn't seem like much information to work with...maybe i'd need to use archimede's principle here as well, to find the weight displacement...but again, I'm not sure how to do that.

Thanks for any help on these problems :)
 
Physics news on Phys.org
HINT: The mass of an object is the product of its density and volume.
 
hmm so mass= density x volume. For the first problem I know the density and that the weight in air is twice that of in water...how do I use that equation if I don't actually know the mass though? Thanks for you're help so far :)
 
Does this help?

Submerged \ Weight = Dry \ Weight - \rho_{water} \ g Volume_{hollow}

Dry \ Weight = \rho_{rock} \ g \ Volume_{rock}
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top