Really lost, pulling ball through water

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Homework Help Overview

The original poster attempts to calculate the force required to pull a copper ball through a fluid at a constant speed, focusing on the effects of drag and weight while ignoring buoyancy. The problem involves concepts from fluid dynamics and forces.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between force, weight, and drag, with some suggesting the need to balance these forces. There are attempts to clarify the calculations involving mass, acceleration, and drag force. Questions arise regarding the correct interpretation of speed units and the application of formulas.

Discussion Status

The discussion includes various calculations and interpretations, with some participants providing guidance on dimensional analysis and the correct use of formulas. There is recognition of potential errors in calculations, particularly regarding unit conversions and assumptions about the problem setup.

Contextual Notes

Participants note discrepancies between calculated results and textbook answers, leading to further exploration of the assumptions made in the problem. The original poster's confusion about the value of gravitational acceleration and the correct speed units is highlighted.

UrbanXrisis
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I need to calculate thr force required to pull a copper ball of radius 2cm upwards through a fluid at the costant speed of 9 cm/s. The drag force is porportional to the speed with porportionality constant 0.950 kg/s. Ignoring the buoyant force, I need to caluate the force.

I fould out the mass=2 .67E-4 kg, after finding copper's density then m=d/v

I think in the end, I will go back to the equation F=ma. So, I need to find acceleration:

a=g-bv/m
b=porportionality constant 0.950 kg/s
v=costant speed of 9 cm/s
m=2 .67E-4 kg

what in the world would g equal? Any help is appreciated
 
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Since the speed is constant, the acceleration is zero.

You need to figure out the force you need to exactly balance the other forces on the ball (its weight plus the drag force). So:
F = weight + drag
 
F=mg+bv/m
=(2 .67E-4 kg)(9.8m/s^2)+(0.950 kg/s)(9m/s)/(2 .67E-4 kg)

like that?
 
UrbanXrisis said:
F=mg+bv/m
=(2 .67E-4 kg)(9.8m/s^2)+(0.950 kg/s)(9m/s)/(2 .67E-4 kg)

like that?

by dimensional analysis, bv/m is in units of acceleration. F=ma, leading me to believe you may want to try bv.
 
F==(2 .67E-4 kg)(9.8m/s^2)+(0.950 kg/s)(9m/s)

The answer I get is 8.8N

However, the book says that it's 3N

What am I doing wrong?
 
you stated the velocity was 9 cm/s not m/s -- that's the only thing that sticks out to me
 
F=(2 .67E-4 kg)(9.8m/s^2)+(0.950 kg/s)(.09m/s)
F=.09N

ackkk where am I wrong?
 
I fould out the mass=2 .67E-4 kg, after finding copper's density then m=d/v

Didn't do the calculation myself but the formula you stated there is wrong.

Remember d=m/v so m=dv
 
thanks! that did the trick!
 
  • #10
I've learned to catch those kind of things because I do them so much myself. Spent forever one night trying to solve a buoyancy problem because I thought it was necessary to solve without the density of an object. When I read the problem again I saw the object was aluminum with a density clearly listed in a table several pages before
 

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