motion due to gravity

[red and white kop!]

When an object falls through a liquid, three forces act on it: its weight, the buoyancy and the resistance of the liquid. Two spheres, of mass 0.5 kg and 1.5 kg respectively, have the same radius, so that they have the same buoyancy of 3.2 newtons, and the same resistance formula, 5v newtons when falling at speed v ms^-1. Both spheres enter the liquid falling vertically at 1 ms^-1. Calculate the terminal speeds of the two spheres, and the acceleration or deceleration when they enter the liquid.
(Here I found: for mass 0.5 kg, terminal speed 0.36 ms^1 and acceleration -6.4 ms^2; for mass 1.5 kg, terminal speed 2.36, acceleration 4.53 (to 3 s.f.))
If the depth of the liquid is 10m, show that the heavier sphere reaches the bottom after a time between 4 and 10 seconds. Find bounds for the time that the lighter sphere takes to reach the bottom.

Ok to starting with the heavier sphere I thought it would be logical to proceed like this:
To find higher bound: distance = 10, u=1, v<2.36, a= (15 – 3.2 – 5v)/1.5 = (11.8 -5v)/1.5, equation s=ut + 0.5at^2, find v in terms of t and find interval for which v<2.36 is true
No problem here, I found t<10
I used a similar method to find the lower bound using s=vt-0.5at^2 and v=1, u>1, but this time the expression had no real roots, so I’m wondering if I’m not using the wrong method altogether, and if my first result was just lucky.
Can someone show me how to work through this?

 

[wjm11]

When an object falls through a liquid, three forces act on it: its weight, the buoyancy and the resistance of the liquid. Two spheres, of mass 0.5 kg and 1.5 kg respectively, have the same radius, so that they have the same buoyancy of 3.2 newtons, and the same resistance formula, 5v newtons when falling at speed v ms^-1. Both spheres enter the liquid falling vertically at 1 ms^-1. Calculate the terminal speeds of the two spheres, and the acceleration or deceleration when they enter the liquid.
(Here I found: for mass 0.5 kg, terminal speed 0.36 ms^1 and acceleration -6.4 ms^2; for mass 1.5 kg, terminal speed 2.36, acceleration 4.53 (to 3 s.f.))
If the depth of the liquid is 10m, show that the heavier sphere reaches the bottom after a time between 4 and 10 seconds. Find bounds for the time that the lighter sphere takes to reach the bottom.

Ok to starting with the heavier sphere I thought it would be logical to proceed like this:
To find higher bound: distance = 10, u=1, v<2.36, a= (15 – 3.2 – 5v)/1.5 = (11.8 -5v)/1.5, equation s=ut + 0.5at^2, find v in terms of t and find interval for which v<2.36 is true
No problem here, I found t<10
I used a similar method to find the lower bound using s=vt-0.5at^2 and v=1, u>1, but this time the expression had no real roots, so I’m wondering if I’m not using the wrong method altogether, and if my first result was just lucky.
Can someone show me how to work through this?

You did not show your work, so I have not checked your numbers. However, I can offer these thoughts:

Force of gravity on mass 1: f = ma => G1 = (.5 kg)(9.81 m/s^2) = -4.905 N (downward)
Buoyancy on mass 1: B1 = 3.2 N (upward)
Total constant force on mass 1: F1 = G1 + B1 = 1.705 N (round to 1.7 if using significant digits)

Conclusion: mass 1 has a net constant upward force; it will sink to some depth, then rise to the surface.

Force of gravity on mass 2: f = ma => G2 = (1.5 kg)(9.81 m/s^2) = -14.715 N (downward)
Buoyancy on mass 2: B1 = 3.2 N (upward)
Total constant force on mass 2: F2 = G2 + B2 = -11.515 N (round to 11.5 if using significant digits)

Conclusion: mass 2 has a net downward constant force; it will reach a negative terminal velocity when drag is equal to the constant forces upon it. Acceleration will be zero at terminal velocity.

Fluid resistance/drag may be assumed(?) to be directly proportional to the square of velocity (was this stated anywhere in the problem?). Zero velocity means zero drag, so you know two points for this (R,v) function: (0,0) and (1,5v). Drag is always opposite the direction of motion, so the sign of R is always opposite that of v. R = kv^2. Start by solving for k.

Total force (TF) will be the sum of the constant force (gravity and buoyancy) and the variable drag force, so TF will vary with time:

TF = F + R

Rearranging f = ma, we see that

a = FT/m = (F + R)/m = F/m + R/m

Velocity and displacement functions can be determined from these considerations (by integrating acceleration) along with initial conditions. However, from the problem statement, it is not clear that exact solutions are required. Maybe just some added, simplifying assumptions to place some bounds on solutions?

Hope that helps.

 

[red and white kop!]

i'm not sure about using integration, its not that complicated, it should just be algebra, not even calc. i'm not sure that your observations helped much but thanks for giving your time, most other forums ignored this question.
the point is i don't know what equations to use seeing that acceleration is not constant, but i'm pretty sure the bounds are to be calculated in function of u=1 and v=terminal speed. but i could be wrong and you might be right about general bounds, but could you elaborate on how to calculate these? also i'm not sure mass 0.5 goes back up after, terminal speed=0.36 ms^-1, i.e. it will continue to move downward.

 

[wjm11]

i'm not sure mass 0.5 goes back up…

Consider two spheres of identical size, one made of balsa wood and the other of steel. What happens when you throw them into water? Both have identical buoyancy (because they displace the same amount of water), but the force of gravity upon them is different. Did you understand what I was doing above when I calculated the forces (F1 and F2) from gravity and buoyancy but neglecting drag? F1 was positive (meaning an upward force) and F2 was negative (meaning a downward force).

The drag force always acts against the direction of motion. If the balls are moving down, drag creates an upward force (a positive force). In the case of the light sphere, both F1 and the drag force R are both positive, i.e., they add together. This means the sphere will slow to zero velocity and then return to the surface. The maximum depth it reaches depends on the initial velocity when it enters the water.

If we neglect drag entirely, what behavior would the light sphere exhibit? The only force on it would be F1 = 1.705 N. It’s initial velocity is –1 m/s, It’s initial position is 0 at the water surface. We want to know when it will reach 10m depth (-10).

From f = ma, we can rearrange to solve for “a”:

a = f/m = 1.705/.5 = 3.41 m/s^2

Displacement equation:

x = x1 +v1(t) + .5(a)(t^2)
-10 = 0 + (-1)(t) + .5(3.41)(t^2)

Rearranging this quadratic, we get:

0 = 1.705t^2 – t + 10

Solving, using the quadratic formula, we get no real solution, meaning the sphere never achieves a depth of 10m.

***

If we check the heavy sphere without any drag in the calculations, we have:

a = f/m = -11.515/1.5 = -7.677 m/s^2

Displacement equation:

x = x1 +v1(t) + .5(a)(t^2)
-10 = 0 + (-1)(t) + .5(-7.677)(t^2)

Rearranging this quadratic, we get:

0 = -3.838t^2 – t + 10

Which yields t = 1.5 s. That is how quickly the sphere would reach 10m if there were no drag (resistance). The sphere would start at –1 m/s and would increase continuously in downward velocity.

Let’s be a little more conservative now and add in the *initial* drag incurred at –1 m/s:

R = 5v = 5(1) = 5 N (positive because the drag force is upward)

Add this to the gravity and buoyancy force:

TF2 = 5 + (-11.515) = -6.515 N

Calculate acceleration:

a = f/m = -6.515/1.5 = -4.343 m/s^2

Displacement equation:

x = x1 +v1(t) + .5(a)(t^2)
-10 = 0 + (-1)(t) + .5(-4.343)(t^2)

Rearranging this quadratic, we get:

0 = -2.172t^2 – t + 10

Solving, we get t = 1.93s. We’ve improved our lower boundary (fastest time) a little from 1.5s to 1.93s to reach a depth of 10m. However, this sphere would still be continuously increasing its downward velocity with our current assumptions.

Let’s find terminal velocity. After rereading your problem statement, I will assume the drag function follows a linear (rather than square) relationship with velocity: R = 5v. When drag is equal and opposite to the combined gravity and buoyancy force, the sphere will have no net force on it, and its velocity will be a constant:

Terminal velocity:

R = G2 + B2 = -11.515 N
5v = -11.515 N
v = -11.515/5 = -2.3 m/s

We can now improve our lower bound again by assuming a constant v = -2.3 m/s over the 10m:

t = d/v = -10/(-2.3) = 4.35s

For an upper bound, assume the velocity never increases beyond –1 m/s:

t = d/v = -10/(-1) = 10s