# Newton's Laws sailboat problem

1. Nov 16, 2003

### Alethia

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.

2. Nov 16, 2003

### Ambitwistor

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(&theta;), Fy = F sin(&theta;). 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.

$${v_f}^2 = {v_i}^2 + 2a\Delta x$$

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. Nov 16, 2003

### 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. =1

I understand how to solve for acceleration using that problem, but how then would I find the frictional force?

4. Nov 16, 2003

### Ambitwistor

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.

The net force is the frictional force, so you just multiply the net acceleration by m.

5. Nov 16, 2003

### Alethia

When you say 6/(sqrt)2 does that mean 6 squared?

m as in mass? There is no mass given...

Last edited: Nov 16, 2003
6. Nov 16, 2003

### Ambitwistor

No, I mean $$6/\sqrt{2}$$.

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.