Really lost, pulling ball through water

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To calculate the force required to pull a copper ball of radius 2 cm upwards through a fluid at a constant speed of 9 cm/s, the drag force must be considered alongside the ball's weight. The mass of the ball is determined to be 2.67E-4 kg using the density of copper. The relevant equation is F = mg + bv, where g is the acceleration due to gravity and b is the proportionality constant for drag. The initial calculations led to confusion regarding units and the correct application of speed, resulting in discrepancies between the calculated force and the expected value. Ultimately, attention to detail in unit conversion and formula application is crucial for accurate results.
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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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