Change Is The Only Constant Thing On The Sea - Part 2
When ships come in to dock where the berths are extremely narrow, the ships needs to be towed in to prevent damage to both the dock and the ship. A metal cable is attached to the front of the ship, looped through a pulley that is hinged at the edge of the dock, and attached to an automated winch.
At low tide, the top of the dock is 6 meters above sea level. The automated winch winds the cable at a constant rate of 10 meters per minute. To 2 decimal places, what is the speed (in meters per minute) at which the boat moves towards the shore when it is 10 meters from the shore?
The answer is 11.66.
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3 solutions
I am concerned about your answer. 11.66m/s is suggesting that pulling an with a force inclined to the horizontal results in a greater force that then applied to the cable. If you resolve the inclined velocity into horizontal and vertical components the horizontal component vb=vcosX where x = tan6/10. Hence vb =8.57
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You are making the incorrect assumption that the rope is traveling in the 'diagonal' tan − 1 1 0 6 direction. However, that's not the actual movement of the rope. The rope length is contracting by 10m/s, and to compensate for that only the horizontal length is changing.
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The rope is the diagonal distance (from the pulley to the attachment on the ship at sea level). The winch pulls the rope in at 10m/s so the diagonal distance contracts by 10m/s.
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@Chris Prowse – Indeed, the diagonal length of the rope contracts by 10m/s. The error in your analysis, is claiming that this must be caused exactly by a vertical component and a horizontal component, in the corresponding ratio. The vertical height remains at 6, and there is no change in the vertical distance. There is only a change in the horizontal component.
In order for the diagonal to contract by 10, the horizontal must contract by more than 10. This is the triangle inequality. Let D be the point on the dock, let X 1 be the initial point on the boat, and let X 2 be the final point on the boat after 1 second. We are given that D X 1 − D X 2 = 1 0 . Hence, by the triangle inequality, since X 1 X 2 + D X 2 > D X 1 hence X 1 X 2 > D X 1 − D X 2 = 1 0
Note that forces are not mentioned at all. The speed of the rope doesn't tell you anything about the forces applied (e.g. the boat's motor could also be running at a muted pace, there could be wave motion, etc).
my explaination about components is off sorry for that here the distance along the cable= sc between boat and dock= sb and height of dock h sc^2=sb^2+h^2 differentiate w.r.t time dt 2,sc.vc=2.sb.vb vb=vc.sc/sb =10.(sqrt of 10^2+6^2)/10 =11.66 my explaination before was not incorrect but i skipped a lot of steps this should be enough
you are talking about force but it was mentioned that the rate of cable being pulled is given 10m/MIN yes the velocity of the boat can be more than that of the cable and indeed the force required to put the boat is more than or equal to the force on boat, but this was never mentioned in the question and there is no need to. your eqn of vb=vcosX is wrong here vb is principle velocity and v is already a component and you took the component of a component to avoid confusion say vb should be divided as vbcosX and vbsinX. vis already a cosine component of v, you cannot take its component again. finally there is no mention of force in the question. It is said that rate of pulling cable is 10m/min and it does not mean that the force is smaller. I want to explain with a diagram but i cant here.
mm
According to my
previous solution
, speed of the boat is
s
i
n
(
θ
)
v
where
v
is the speed of pulling the cable. So, we get
d
t
d
x
=
s
i
n
(
θ
)
v
=
6
2
+
1
0
2
1
0
1
0
=
1
1
.
6
6
.
Yours is incorrect too. It's a decreasing cable and therefore the rate is negative -11.66.
I did it by the chain rule, let the cable from the winch to the boat be h, the length from the boat to the dock be x, dh/dt = 10 which is given, x= sqrt(h^2-36) by pythagoras so dx/dh=h/sqrt(h^2-36). dx/dt = dh/dt x dx/dh which is 10h/(sqrt(h^2-36). when x=10 h=sqrt(136) by pythagoras, sub it in and voila, 11.66.
grrrrg! I can't answer this.😠
consider velocity of cable v velocity of boat Vb now from the diagram, v=Vb.cosX (consider X as angle between Vb and v, cosX= 10/(sqrt of 10^2+6^2)) Vb= v/cosX = 10/(10/sqrt of 136) = 11.66