Force and acceleration given position equations. Find magnitude and angles

In summary, A 0.42 kg particle moves in an xy plane according to x(t) = - 19 + 2 t - 2 t3 and y(t) = 25 + 8 t - 7 t2, with x and y in meters and t in seconds. At t = 0.6 s, the magnitude of the net force on the particle is 6.61 N. The angle of the net force within the (-180°, 180°] interval relative to the positive direction of the x axis is 62.78 deg. The angle of the particle's direction of travel is also 62.78 deg.
  • #1
pizzie
4
0

Homework Statement


A 0.42 kg particle moves in an xy plane according to x(t) = - 19 + 2 t - 2 t3 and y(t) = 25 + 8 t - 7 t2, with x and y in meters and t in seconds. At t = 0.6 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


Homework Equations


Fx = ma(x)
Fy = ma(y)

The Attempt at a Solution


I got the correct answer for part A. I found the second derivative for the x(t) and y(t)
x"(t) = -12t
y"(t) = -14

to find Fx i used the equation Fx = ma(x) = (0.42)(-12)(0.6) = -3.024
to find Fy i used Fy = ma(y) = (0.42)(-14) = -5.88

i then squared both of those values, added them together, and took the square root to get the magnitude, which is 6.61. this is the right answer.

For part B i took the inverse tangent of Fy/Fx = arctan(-5.88/-3.024) and i got 62.78 deg, but this is the wrong answer.

For C i took the inverse tangent of the first derivative equations for velocity.
Vx = 2 - 6(0.6)^2 = -0.16
Vy = 8 - 14(0.6) = -0.4
then i did arctan(-0.4/-0.16), but this is also wrong.

i'm not sure what I'm doing wrong for B and C. any help would be appreciated. thank you
 
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  • #2
Both the cosine and the sine of the angle of the force are negative. In which quadrant is that angle? The same is with the velocity. Both components are negative, what is its angle with respect to the positive x axis?

ehild
 
  • #3
thank you! :) :)
 

1. What is the relationship between force and acceleration?

The relationship between force and acceleration is described by Newton's second law of motion, which states that the force applied to an object is directly proportional to its mass and acceleration. This means that the greater the force applied to an object, the greater its acceleration will be.

2. What is a position equation?

A position equation is a mathematical expression that describes the position of an object at a given time. It takes into account the initial position, velocity, and acceleration of the object and can be used to calculate its position at any point in time.

3. How do you find the magnitude of force and acceleration using position equations?

To find the magnitude of force and acceleration using position equations, you can use the formula F=ma, where F is the force, m is the mass of the object, and a is the acceleration. Substitute the values from the given position equation to calculate the magnitude of force and acceleration.

4. What do the angles in a position equation represent?

The angles in a position equation represent the direction of the force and acceleration vectors. These angles can be used to determine the direction in which the object is moving and the direction of the force acting on the object.

5. How can position equations be applied in real-world scenarios?

Position equations are commonly used in physics to describe the motion of objects. They can be applied in real-world scenarios such as calculating the trajectory of a projectile, determining the forces acting on a moving vehicle, or predicting the position of planets in the solar system.

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