Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Finding downward force on immersed object
Reply to thread
Message
[QUOTE="kuruman, post: 6867545, member: 192687"] A clue to that is the sloppy derivation, item 4, quoted by [USER=731016]@Callumnc1[/USER] in post #43. We have [B]Step 1[/B] ##\rho_{\text{object}}=\dfrac{m}{V}.## This is OK. [B]Step 2[/B] ##\dfrac{m}{V_{\text{object}}}=-\dfrac{m \rho_{\text{water}} g}{B}.## To see where this comes from, note that ##B=\rho_{\text{water}} V_{\text{object}} g \implies V_{\text{object}}=\dfrac{B}{\rho_{\text{water}} g}.## Replacing that in the denominator gives ##\dfrac{m}{V_{\text{object}}}= \dfrac{m \rho_{\text{water}} g}{B}.## There should be no negative sign because that would make the density on the left hand side negative. The buoyant force ##B## cannot be negative because it is equal to ##\rho_{\text{water}} V_{\text{object}} g##, which is positive. [B]Step 3[/B] ##\dfrac{m\rho_{\text{water}} g}{B}=\dfrac{F_g}{F_i-F_g}\rho_{\text{water}}.## (I dropped the negative signs that appear on both sides of the equation.) Here the buoyant force in the denominator has been replaced by the difference between the weight in air and the mysterious ##F_i##. Note that $$\frac{F_g}{F_i-F_g}\rho_{\text{water}}=\left( \frac{F_i-F_g}{F_g} \right)^{-1}\rho_{\text{water}}=\left( \frac{F_i}{F_g}-1 \right)^{-1}\rho_{\text{water}}.$$ This is the [B]negative[/B] of what the professor expected to get and explains the negative sign in Step 2. It is a fudge factor to make the answer come right. When replacing the buoyant force with a difference, that difference should have been ##F_g-F_i## where ##F_i## is the reading of the force gauge with the mass under water, also the tension in the string from which the mass is hanging, also the force that the net exerts on the mass. Where this comes from is shown in post #24, equations (1) and (2). To [USER=731016]@Callumnc1[/USER] : Look at my derivation in post #38 and forget the other mess. ##F_i## is the force that you recorded with the masses under water. You don't have to calculate it because you measured it. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Finding downward force on immersed object
Back
Top