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Newton's Laws sailboat problem

  • Thread starter Alethia
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  • #1
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Okay, in my class we've been learning basic problems solved by applying Newton's laws. I understand the general idea, but I get confused with problems like the first one. How would I solve it taking both forces on it into account? On the second problem, I don't know how to solve it because it gives you velocity and none of the formulas have velocity in it. Do I have to refer to other formula's first and then convert? Any explanation or help would be very much appreciated. Please provide a step-by-step guide so that I can teach myself. THANK YOU.

1) A sailboat with a mass of 2.0x10^3kg experiences a tidal force of 3.0x10^3N directed to the east and a wind force against its sails with a magnitude of 6.0x10^3N directed towards the northwest (45 degrees North of West). What is the magnitude of the resultant acceleration of the boat?

2) A crate is carried in a pickup truck traveling horizontaklly at 15.0m/s. The truck applies the brakes for a distance of 28.7m while stopping with uniform acceleration. What is the coefficient of static friction between the crate and the truck bed if the crate does not slide?

Thanks again for any help, and sorry to inconvienence anyone.
 

Answers and Replies

  • #2
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Originally posted by Alethia
I understand the general idea, but I get confused with problems like the first one. How would I solve it taking both forces on it into account?
You have to resolve the vectors into components. If you take 'x' to be east and 'y' to be north, then the tidal force is 3.0x10^3 x, and the wind force is 6.0/sqrt(2)x10^3 x + 6.0/sqrt(2)x10^3 y.

(To resolve the wind vector into components, you note that it makes an angle of 45 degrees with respect to the x axis, so the components of the vector are Fx = F cos(θ), Fy = F sin(θ). Here we use that cos(45)=sin(45) = 1/sqrt(2).)

The net force is thus (3+6/sqrt(2))x10^3 x + 6/sqrt(2)x10^3 y; use the Pythagorean theorem to find its magnitude.


On the second problem, I don't know how to solve it because it gives you velocity and none of the formulas have velocity in it.
What about this one?

[tex]
{v_f}^2 = {v_i}^2 + 2a\Delta x
[/tex]

It's an underrated but very useful kinematics formula. You can solve for the net acceleration, which is due to the frictional force, and from that determine the coefficient of friction.
 
  • #3
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I don't understand what you did for the first problem. By following your explanation I got the answer of 268.3 which is not the correct answer. =1

I understand how to solve for acceleration using that problem, but how then would I find the frictional force?
 
  • #4
841
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Originally posted by Alethia
I don't understand what you did for the first problem. By following your explanation I got the answer of 268.3 which is not the correct answer.
Ooops, I got the sign wrong. The wind force is
-6.0/sqrt(2)x10^3 x + 6.0/sqrt(2)x10^3 y (since it's northwest, not northeast).

You should obtain an acceleration of 2.21 m/s2.


I understand how to solve for acceleration using that problem, but how then would I find the frictional force?
The net force is the frictional force, so you just multiply the net acceleration by m.
 
  • #5
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When you say 6/(sqrt)2 does that mean 6 squared?

Originally posted by Ambitwistor
The net force is the frictional force, so you just multiply the net acceleration by m.
m as in mass? There is no mass given...
 
Last edited:
  • #6
841
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Originally posted by Alethia
When you say 6/(sqrt)2 does that mean 6 squared?
No, I mean [tex]6/\sqrt{2}[/tex].


m as in mass? There is no mass given...
Yes. Multiply by mass anyway, and when you go to calculate the coefficient of friction from the force of friction, you'll find it cancels out of the equation so you don't need to know its value.
 

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