# Homework Help: Particle of certain mass moving in an xy plane

1. Oct 9, 2011

### eximius

1. The problem statement, all variables and given/known data

A 0.39 kg particle moves in an xy plane according to x(t) = - 13 + 1 t - 6 t^3 and y(t) = 28 + 5 t - 11 t^2, with x and y in meters and t in seconds. At t = 1.0 s, what are (a) the magnitude and (b) the angle (within (-180°, 180°] interval relative to the positive direction of the x axis) of the net force on the particle, and (c) what is the angle of the particle's direction of travel?

2. Relevant equations

F(y) = ma(y)
F(x) = ma(x)

3. The attempt at a solution

Well I've got the magnitude of the force correct as 16.45N after finding the second derivative of x and y and then using Pythagoras's theorem.

What I'm struggling with is part B and C. From what I understand both answers should be the same, as the direction of Fnet should be the direction of travel.

I attempted to calculate the angle between the force and the y axis. I used sin^-1 theta = 22/28.43. This gave me a value for theta of 50.71 degrees. I then added 90 degrees to this to get the angle relative to the positive x axis. Getting an answer of 140.71 degrees for both B and C. But this is apparently wrong.

Where did I go wrong?

Last edited: Oct 9, 2011
2. Oct 9, 2011

### cepheid

Staff Emeritus
The y-component and the x-component of the force, together with the total force vector, form a right triangle. The angle you want has the y-component as the side opposite to it, and the x-component as the side adjacent to it. Therefore:

θ = tan-1(Fy/Fx)

3. Oct 9, 2011

### cepheid

Staff Emeritus
Also, for part C, you want the angle of the position vector, not the force vector. So the answers to part B and C will NOT be the same.

4. Oct 9, 2011

### eximius

Therefore the answer to B will be:

θ = tan-1(8.58/14.04) = 31.43 degrees
Relative to positive x axis = 180 - θ = 148.57 degrees

And the answer to C will be:

θ = tan-1(22/18) = 50.71 degrees
Relative to positive x axis = θ + 90 = 140.71 degrees

Is that right? And thanks for the help, it's much appreciated.

5. Oct 9, 2011

### cepheid

Staff Emeritus
I don't understand why you're doing these weird shifts. Draw a picture. Assuming that you're taking up to the positive y-direction and rightwards to be the positive x-direction, then the original angles you get (31.43 and 50.71) are fine (i.e. they are already measured from the positive x-axis). They lie within the specified ranges and there is no need to apply these offsets to them.

Note: I haven't checked your arithmetic. What I'm saying is that assuming that 31.43 and 50.71 are actually correct values in the first place, then there is no need to do anything else to them.

6. Oct 10, 2011

### eximius

I used the 2 equations to calculate the position of the particle at t = 1, I get (-18,22). This shows that the particle is travelling North West. Doesn't the negative value of x indicate that I need to resolve the angles to the 2nd quadrant?