Dynamics Laminar Flow Problem

[dinkadink]

Problem:
Two horses pull a barge from rest in a canal filled with a viscous fluid that provides laminar resistance (FR = –kv). The two horses walk on each side of the canal so that their net force is applied exactly forward. The barge has a mass of 3000 kg and the donkeys pull forward with a constant net forward force of 300 Newtons.

a. If the maximum attainable speed by the barge is 2.5 m/s, what is the resistive coefficient, k (include units)?
b. How long does it take the barge to reach a speed of 1 m/s?

Once the barge reaches 2.5 m/s, at a certain time (call it t = 0) the horses instantaneously stop pulling, and the barge is allowed to drift forward.

c. How long does it take for the barge to slow to 1 m/s?
d. How far does the barge drift before coming to rest?
e. In principle, how long does it take for the barge to come to rest? Briefly explain.
f. Estimate how long it takes for the bar to be, for all practical purposes, at rest. Discuss how you decided on this estimate.

Equations:
F=-kv

Attempt at a solution:
I know how to do the integrations of the laminar flow F=-kv. Other than that I really don't know how to do this.

 

[Nate810]

Please help!a. k = -300 N / (2.5 m/s) = -120 N/m/sb. t = 1/k * ln(v1/v2) = 1/-120 N/m/s * ln(1/2.5) = 2.30 seconds c. t = 1/k * ln(v2/v1) = 1/-120 N/m/s * ln(2.5/1) = 2.30 seconds d. d = v2*t = 2.5 m/s * 2.30 s = 5.75 me. It takes an infinite amount of time for the barge to come to rest, since the force of viscous friction is proportional to the velocity and decreases as the velocity approaches zero. f. This is dependent on the magnitude of the resistive coefficient, k. An estimate of the time it would take for the barge to be practically at rest can be calculated by assuming that the resistance force is equal to the net forward force at a certain velocity. For example, if we assume that the resistance force is equal to the net forward force at 0.10 m/s, then the time it would take for the barge to be practically at rest can be calculated by solving the following equation: Fnet = Fres = -kv = 300 N. Solving this equation for v gives us v = 300 N/-120 N/m/s = 2.50 m/s. The time it would take for the barge to reach this velocity can be found by using the equation t = 1/k * ln(v1/v2) which gives us t = 1/-120 N/m/s * ln(0.10/2.50) = 1.94 seconds. Therefore, it would take approximately 1.94 seconds for the barge to be practically at rest.

 

[Moetasim]

I would approach this problem by first identifying the key elements and assumptions. The problem involves two horses pulling a barge in a canal filled with a viscous fluid that provides laminar resistance. The barge has a mass of 3000 kg and is pulled with a constant net forward force of 300 Newtons. We are given the maximum attainable speed of the barge (2.5 m/s) and asked to find the resistive coefficient, k. We are also asked to calculate the time it takes for the barge to reach a speed of 1 m/s, and to determine how long it takes for the barge to slow to 1 m/s and come to a complete stop.

To solve this problem, we can use the equation F=-kv, where F is the force applied by the horses, k is the resistive coefficient, and v is the velocity of the barge. We can rearrange this equation to solve for k: k = -F/v. Substituting the given values, we can calculate k as follows:

a. k = -300 N / 2.5 m/s = -120 kg/s

b. To calculate the time it takes for the barge to reach a speed of 1 m/s, we can use the equation v = at, where a is the acceleration and t is the time. Since the barge is pulled with a constant force, we can use Newton's second law (F=ma) to find the acceleration: a = F/m = 300 N / 3000 kg = 0.1 m/s^2. Substituting this value into the equation, we get: t = v/a = 1 m/s / 0.1 m/s^2 = 10 seconds.

c. To calculate the time it takes for the barge to slow to 1 m/s, we can use the same equation (v = at), but with a negative acceleration since the barge is slowing down. Using the acceleration calculated in part b, we get: t = v/a = 1 m/s / (-0.1 m/s^2) = -10 seconds. However, since time cannot be negative, we can assume that the barge takes 10 seconds to slow down to 1 m/s.

d. To calculate the distance the barge drifts before coming to rest, we can use the equation v^2

 

[ogb p]

I would approach this problem by first analyzing the given information and identifying the relevant equations and concepts. From the problem, we know that the barge is being pulled by two horses with a constant forward force of 300 Newtons and that the barge experiences a resistive force due to the viscous fluid, given by F = -kv. We also know that the maximum attainable speed of the barge is 2.5 m/s.

a. To find the resistive coefficient, we can use the equation F = ma, where F is the net force on the barge, m is the mass of the barge, and a is its acceleration. Since we are given the maximum speed, we can assume that at this speed, the net force is zero. Therefore, we can set F = 0 and solve for k: 0 = -k(3000 kg)(2.5 m/s). This gives us a value of k = 0.04 Ns/m.

b. To find the time it takes for the barge to reach a speed of 1 m/s, we can use the equation v = u + at, where v is the final speed, u is the initial speed (which is 0 in this case), a is the acceleration (which is given by F/m = -kv/m), and t is the time. Plugging in the values, we get 1 m/s = (0 m/s) + (-0.04 Ns/m)(t), which gives us a time of t = 25 seconds.

c. When the horses stop pulling, the net force on the barge becomes the resistive force, which will cause the barge to decelerate. We can use the same equation as in part b to find the time it takes for the barge to slow to a speed of 1 m/s, but this time we will use a positive value for the acceleration since it is acting in the opposite direction. This gives us a time of t = 25 seconds.

d. To find the distance the barge drifts before coming to rest, we can use the equation x = ut + 1/2at^2, where x is the distance, u is the initial speed, a is the acceleration, and t is the time. Since the barge is starting from a speed of 2.5 m/s and coming to rest, we can use