Physics Help Forum Finding density of an object with the weight and density of fluid?

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Apr 26th 2019, 06:15 AM #1 Junior Member   Join Date: Apr 2019 Posts: 23 Finding density of an object with the weight and density of fluid? Hello, I am a bit stuck with a question "A large container of water initially reads 0.485 kg. A ball is hung in the air from another set of scales which initially read 0.112 kg. The ball is lowered until it is fully submerged. The readings on both scales are now 0.496 kg and 0.101 kg respectively. What is the density of the ball?" My working so far is here: https://imgur.com/a/sINtKX8 I have the volume of the fluid displaced and the density of the water which is provided, (p=1000kg m-3) but I'm unsure how to get the volume of the object in order to work out the density with the equation written in the far right (Vd/Vo=po/pf). Would anyone know what the next step is? Thank you -I
 Apr 26th 2019, 07:26 AM #2 Senior Member     Join Date: Jun 2016 Location: England Posts: 1,000 Note that the change in reading of both scales is 0.011 kg The volume of water displaced must equal the volume of the ball. __________________ ~\o/~
 Apr 26th 2019, 07:43 AM #3 Senior Member   Join Date: Jun 2010 Location: NC Posts: 418 Operant Physical Law: Newton's 2nd Law... Use It! Hi, Newton's 2nd Law relates the forces you study. So start by selectng a system then writing the 2nd for it. Here are three similar problems that show the method. Volume of a Sculpture Civil War Memorial | THERMO Spoken Here! First Equestrian Ascent Good Luck... TSH
 Apr 26th 2019, 08:07 AM #4 Junior Member   Join Date: Apr 2019 Posts: 23 Hello, this must mean that 1 = pobj/1000 pobj = 1 x 1000 =1000 But this isn't the right answer. Is something wrong with my rearranging here maybe? And thank you for your reply, as always.
 Apr 26th 2019, 08:48 AM #5 Senior Member     Join Date: Jun 2016 Location: England Posts: 1,000 The Ball pushes a volume of water (equal to the balls own volume) out of the way, however the displaced water "pushes back" The Force pushing back is equal to the Weight of water displaced. Be aware of the distinction between Weight (which is a force) and Mass (which isn't). __________________ ~\o/~
 Apr 26th 2019, 09:01 AM #6 Junior Member   Join Date: Apr 2019 Posts: 23 So the force pushing back is 0.11N, because 0.011 x 10 = 0.11 (using 10 as gravity instead of 9.8). So 0.11/ (1000 x 10) = 1.1x10^-5 kgm-3 This is the volume, and since the volume of the object is equal, 1.1x10^-5 divided by itself is one, which leads me back to the same answer.
 Apr 26th 2019, 10:03 AM #7 Senior Member     Join Date: Jun 2016 Location: England Posts: 1,000 The change in Force measured by the scales is 0.11 Newtons (using g=10) This implies a displacement of 0.011 kg of water, This 0.011 kg of water is "trying to return to its original position" (Ok this is anthropomorphizing a puddle of water, but sometimes it helps in the visualisation) And pushing back against the ball with all its 0.11 Newtons of Weight. 1 kg of water has a volume of 1 litre (at standard room temperature and pressure). Thus 0.011 kg of water has a volume of 0.011 litres Thus the Ball has a volume of 0.011 litres. The Ball has a mass of 0.112 kg. Thus it is a very dense ball... __________________ ~\o/~
 Apr 27th 2019, 04:42 AM #8 Junior Member   Join Date: Apr 2019 Posts: 23 IT IS A DENSE BALL WHICH MEANS p=m/V And 0.112/1.1x10^-5 = 10181 = 10200kgm-3 IT'S VERY DENSE AND ALSO THE ANSWER THANK YOUUU! i cannot express my gratitude. Also, anthropromorphising not alive things is always useful, in every context, ever! Thanks again! Last edited by astupidummydumstupid; Apr 27th 2019 at 04:43 AM. Reason: changed ever to every

 Tags density, finding, fluid, object, weight

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