Tension in string of object suspended in liquid

Click For Summary
The discussion focuses on calculating the tension in a string supporting an aluminum object suspended in ethyl alcohol. The key equations used include the balance of forces, where tension (T) equals the weight of the object minus the buoyant force (Fb). The calculations reveal that the mass of the aluminum object is 0.27 kg, while the buoyant force from the ethyl alcohol is 0.079 kg. The effective mass of the block is determined to be 0.191 kg, leading to the final weight calculation. Emphasis is placed on the importance of including units in calculations to avoid errors.
javashackgirl
Messages
2
Reaction score
0

Homework Statement



an aluminum object with volume 100 cm^3 and density 2700 kg/m^3 is suspended in ethyl alcohol (density 790) by a string. what's the tension in the string?



Homework Equations



Fy = Fb + T - mg
T = mg - Fb
Fb = p_f * v_f * g = p_o * v_o * g

The Attempt at a Solution



=(p_o - p_f)p_f*v_o*g

=(2700 - 790)(790)(1*10^-4)(9.81)

Thanks!
 
Physics news on Phys.org
Not quite, check the units.

It might be simpler if you worked in some numbers.
Tension = weight down - boyancy up
mass = density * volume = 2700 kg/m^3 * 0.0001 m^3 = 0.27 kg
boyancy = weight of fluid = 790 kg/m^3 * 0.0001m^3 = 0.079 kg

So effective mass of block (0.27-0.079) = 0.191kg,
weight = 0.1919kg * 9.81 Newtons
 
i figured out what i did wrong. your way was much easier, thanks :)
 
Useful tip always put the units in your calculation or rearrangment of equations - it makes it much easier to spot any little 'opps'

Eg. to get mass from density.
mass = density * volume
kg = kg m^-3 * m^3 = kg !
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 12 ·
Replies
12
Views
1K
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 37 ·
2
Replies
37
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
25K
  • · Replies 19 ·
Replies
19
Views
3K