Really lost, pulling ball through water

In summary, the conversation discusses the process of calculating the force required to pull a copper ball of radius 2cm upwards through a fluid at a constant speed of 9 cm/s. The drag force is proportional to the speed with a proportionality constant of 0.950 kg/s. The mass of the ball is found to be 2 .67E-4 kg and the formula F=ma is used to find the acceleration needed. The conversation also touches on the importance of correctly interpreting units and using the correct formula to solve the problem.
  • #1
UrbanXrisis
1,196
1
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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  • #2
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
 
  • #3
F=mg+bv/m
=(2 .67E-4 kg)(9.8m/s^2)+(0.950 kg/s)(9m/s)/(2 .67E-4 kg)

like that?
 
  • #4
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.
 
  • #5
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?
 
  • #6
you stated the velocity was 9 cm/s not m/s -- that's the only thing that sticks out to me
 
  • #7
F=(2 .67E-4 kg)(9.8m/s^2)+(0.950 kg/s)(.09m/s)
F=.09N

ackkk where am I wrong?
 
  • #8
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
 
  • #9
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 :grumpy:
 

1. How does the density of water affect the movement of a ball?

The density of water affects the movement of a ball by creating resistance or drag as the ball moves through the water. The higher the density of water, the more drag it will create, making it harder for the ball to move through the water.

2. Why does a ball move slower in water compared to air?

A ball moves slower in water compared to air because water has a higher density than air, which creates more resistance or drag on the ball. This drag force acts in the opposite direction of the ball's movement, slowing it down.

3. How does the size and shape of a ball affect its movement through water?

The size and shape of a ball can greatly affect its movement through water. A larger and more streamlined ball will experience less drag, allowing it to move faster through the water. A smaller or more irregularly shaped ball will experience more drag and move slower.

4. What factors can influence the speed of a ball moving through water?

The speed of a ball moving through water can be influenced by various factors such as the density and temperature of the water, the size and shape of the ball, and any external forces acting on the ball, such as currents or waves.

5. How can the movement of a ball through water be calculated or predicted?

The movement of a ball through water can be calculated using various formulas and equations, such as the drag equation and the buoyancy equation. Factors such as the ball's size, shape, and velocity, as well as the properties of the water, must be taken into account for an accurate calculation or prediction.

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