3
$\begingroup$

We are doing problems involving boats sailing from one shore to the other. Usually, in such problems we are told the velocity of the boat with respect to the water and the angle at which the boat sets out. We are then told the speed and direction of the stream (from ground).

My question is why do we assume that the flow of water effects all boat paths equally. For example, if one boat is much more massive than the other and both have the same velocity with respect to water and both set out at the same angle, how is it that the stream deviates the path of both by the same amount? Shouldn't the more massive one have deviated less? But we don't include mass in any of our calculations.

What am I missing?

$\endgroup$
  • $\begingroup$ You are correct that the simplified model of a boat isn't realistic at all. But, it is used to demonstrate certain mathematics with a quasi visual way that is more helpful than harmful. $\endgroup$ – ja72 Oct 9 '19 at 1:14
  • 2
    $\begingroup$ @ja72 What's unrealistic? If you were given velocity of a boat with respect to water, and water with respect to ground, you should be able to determine the velocity with respect to ground, regardless of the boat. That's already factored into the hypothetical information. $\endgroup$ – JMac Oct 9 '19 at 1:23
  • $\begingroup$ @JMac - the velocity of the boat is not prescribed. In real life, velocity is a result of accelerations acting over time. It is true that the boat only cares about the relative velocity difference between itself and the water underneath, but the response is subject to dynamics and the standard problem of this kind only deals with the kinematics of the problem. $\endgroup$ – ja72 Oct 9 '19 at 13:20
  • 2
    $\begingroup$ @ja72 I don’t see how that makes it unrealistic. We are directly told how the boat is moving with respect to water. There’s nothing unrealistic about that. It wouldn’t be unrealistic to say a man on a train moves at some velocity relative to the train, even though human walking is also a complex dynamic process. The information given means we don’t have to consider any of that, we are told the consequences already. $\endgroup$ – JMac Oct 9 '19 at 13:32
3
$\begingroup$

You are given the velocity of the boat with respect to the water. You are also given the movement of water with respect to the ground.

You don't assume that the water affects all boat paths equally. If you are told the velocity of the boat with respect to the water, you know how much the water affects the movement of the boat, because you are already given the speed the boat moves relative to the water.

$\endgroup$
  • $\begingroup$ For e.g. we were told that the boats move 16 km/h relative to the water. Now suppose one is light and the other is massive; both head north in a certain stream. Where has mass come in in this information? $\endgroup$ – Sal_99 Oct 8 '19 at 16:28
  • $\begingroup$ @Sal_99 It doesn’t. Mass would let you determine how much energy it takes to move at that speed. Since we don’t care how much energy that takes, we don’t have to think about it. The question tells you the speed, mass doesn’t change the speed because you already know it. $\endgroup$ – JMac Oct 8 '19 at 16:50
  • $\begingroup$ yes I get your point but the deviation in the path created( i.e.the boats were heading north but end up somewhere else ) is assumed to be the same, regardless of mass here, no? That is what I was trying to ask about. $\endgroup$ – Sal_99 Oct 8 '19 at 16:57
  • $\begingroup$ @Sal_99 It’s the same because the question tells you it’s the same. There’s no assumption required, you’re told the velocity irrespective of mass, there is no assuming to be done for that. $\endgroup$ – JMac Oct 8 '19 at 17:14
  • 1
    $\begingroup$ @Sal_99 The question never really mentions how the boat achieves those velocities with the water, and it doesn't matter, we are given enough information. What is there to justify? Velocity of the boat with respect to the water is any hypothetical value they want it to be. Angle at which the boat sets out is also hypothetical and up to them. The velocity of the stream with respect to ground is also hypothetical value up to them. But with those three values you can absolutely determine the hypothetical velocity with respect to the ground. It's a justified property of additive velocities. $\endgroup$ – JMac Oct 9 '19 at 1:21
1
$\begingroup$

In these questions, they assume that the stream is much much bigger than the boat. Hence the boat will go along with the stream regardless of its size. You don not have to assume anything as such questions will never deal with numerous boats each having different masses.

$\endgroup$
  • 2
    $\begingroup$ They don't even have to assume that. They provide the velocity of the boat relative to the stream and the velocity of the stream relative to the ground. With the information given, you can determine the velocity with respect to the ground without assuming anything. It's just a fact about the hypothetical. $\endgroup$ – JMac Oct 8 '19 at 17:39
0
$\begingroup$

These problems tend to oversimplify the system being analyzed. So, it is certainly not always true that boats of different masses are affected equally.

But that isn't to say it can't. If we want both boats to move at same velocity then we need the net force to equal zero, which requires that the force of the water pushing on the boat is equal in magnitude to the drag force. Assuming the dominate drag force in the water is from the quadratic term, we have:

$$F_{water} = -F_{drag}$$ $$P_{water} *(Area) = cv^2$$

where the area is the cross sectional area, P is the pressure supplied by the water, and c is a constant that happens to be proportional to the cross sectional area. The velocity for a given object is then:

$$v = \sqrt{\frac {P_{water}(Area)}{c}}$$

Since the area and c are both proportional to the cross sectional area, and an increase in mass results in a change in cross sectional area (Archimedes Principle), then I imagine it is possible the speed would remain unchanged for different masses. It all depends on how exactly c changes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.