Density of bar submerged in water

AI Thread Summary
The discussion focuses on calculating the upthrust and density of a bar submerged in water, where the bar weighs 1.75 N in air and 1.4 N in water. The upthrust is correctly calculated as 0.35 N, representing the buoyant force. However, there is confusion regarding the calculation of the bar's density, as the volume derived from the weight of water displaced is incorrectly computed. It is emphasized that density should be calculated using mass rather than weight, and Archimedes' Principle must be applied correctly to determine the buoyant force based on the weight of the water displaced. Accurate calculations are crucial for resolving the apparent contradiction regarding the bar's ability to float.
Roodles01
Messages
125
Reaction score
0

Homework Statement


A bar suspended in air weighs 1.75 N
The same bar weigs only 1.4 N when suspended in water.
Calculate;
a. upthrust
b. density of the bar

Homework Equations


D = m/v
F = m*a

The Attempt at a Solution


a[/B].
Upthrust is 1.75 N - 1.4 N = 0.35 N

b.
Assuming density of water is 1x103 kg m-3
and weight of bar displaces that amount of water.
So
volume 1.75 N of water = 1.75 N / (1x103 * g) = 1.78 m3

Density of bar = m/v = 1.75 N / 1.78 m3 = 0.98 N m-3

which has to be wrong, surely, or the bar would float . . .
Help!
Thank you.
 
Physics news on Phys.org
Remember the difference between weight and mass.
 
Roodles01 said:

Homework Statement


A bar suspended in air weighs 1.75 N
The same bar weigs only 1.4 N when suspended in water.
Calculate;
a. upthrust
b. density of the bar

Homework Equations


D = m/v
F = m*a

The Attempt at a Solution


a[/B].
Upthrust is 1.75 N - 1.4 N = 0.35 N

b.
Assuming density of water is 1x103 kg m-3
and weight of bar displaces that amount of water.
So
volume 1.75 N of water = 1.75 N / (1x103 * g) = 1.78 m3

It's not clear how you got 1.75 N / (9810) = 1.78 m3

Your calculator must have suffered a nervous breakdown.

Remember, a cubic meter of water weighs a tonne (1000 kg), literally.

Density of bar = m/v = 1.75 N / 1.78 m3 = 0.98 N m-3

which has to be wrong, surely, or the bar would float . . .
Help!
Thank you.

As Dr. Courtney pointed out, density is mass divided by volume, not weight divided by volume. You can't use Newtons and kilograms interchangeably.
 
Roodles01 said:

Homework Statement


A bar suspended in air weighs 1.75 N
The same bar weigs only 1.4 N when suspended in water.
Calculate;
a. upthrust
b. density of the bar

Homework Equations


D = m/v
F = m*a

The Attempt at a Solution


a[/B].
Upthrust is 1.75 N - 1.4 N = 0.35 N
yes, and this called the buoyant force
b.
Assuming density of water is 1x103 kg m-3
and weight of bar displaces that amount of water
So
volume 1.75 N of water = 1.75 N / (1x103 * g) = 1.78 m3
you are not applying Archimedes Principle correctly. The upward thrust or buoyant force is equal to the weight of water displaced. That weight of displaced water is not 1.75 N.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top