Magnitude and direction of the net force vector

[XwakeriderX]

Homework Statement

The position r as a function of time of a 15.0 kg mass, moving
in the x-y plane, is given by r(t) = (6.00t^3 x + 1.20t^4 y) meters where x and y are unit vectors. At t = 5.00 s, find the magnitude and direction of the net force vector
which is causing this acceleration

Homework Equations

a=(Ax)x + (Ay)y +(Az)z
and each variable after the () has a hat on them

The Attempt at a Solution

I don't know how to start the problem : /

 

[PhanthomJay]

Are you familiar with calculus and acceleration as a function of displacement?

 

[XwakeriderX]

Yes, this is going to sound strange but I am really good at calculus but whhen it comes to physics I am having troubles setting things up into the right format and i have no examples at all to follow. If i could just see a graph of what I am looking at all would make sense

 

[PhanthomJay]

In the Physics motion equations, the instantaneous velocity is the first derivative of the displacement vector with respect to time, and the acceleration vector is the 1st derivative of the velocity vector with respect to time. But then you need to know and apply which law to solve for the net force causing the acceleration?

 

[XwakeriderX]

Im sorry i am really trying, I see the formula i know how to solve that..i just don't understand what variables go where. r(t) = (6.00t^3 x + 1.20t^4 y) leaves me very confused. i think its the powers that are throwing me off i can't find any examples like this

 

[PhanthomJay]

Im sorry i am really trying, I see the formula i know how to solve that..i just don't understand what variables go where. r(t) = (6.00t^3 x + 1.20t^4 y) leaves me very confused. i think its the powers that are throwing me off i can't find any examples like this

The derivative of a sum is the sum of the derivatives of each term. Look at the x direction and y direction, separately. Those x and y's in the equation are confusing. they indicate the component directions of the displacemnt. I'll start you off with dr/dt in the x direction: d/dt(6t^3) =18t^2. Continue...

 

[XwakeriderX]

(thank you so much for being patient)
dr/dt in y direction d/dt(1.20t^4)=4.8t^3
so now i have the derivative of the x direction and y direction. Where can i go from here? when will the mass and seconds come in play?

 

[PhanthomJay]

That's the first derivative...now differentiate again to find the acceleration components...and their magnitude and direction at t=5 seconds...

 

[XwakeriderX]

Okay! so now i have 14.4t^2 +36t has my acceleration equation! so my acceleration at 5 seconds is 540 m/s ?!? now how can i use this to find the magnitude and direction! finally I am getting a breakthough(hopefully haha)

 

[PhanthomJay]

Okay! so now i have 14.4t^2 +36t has my acceleration equation! so my acceleration at 5 seconds is 540 m/s ?!? now how can i use this to find the magnitude and direction! finally I am getting a breakthough(hopefully haha)

Not quite. the acceleration is 14.4t^2 in the x direction and 36t in the ydirection. You have to look in each direction separately. That's what the 'hat' indicated in the original equation...x_hat is the displacement in the x direction, and y_hat is the displacement in the y direction. So you have to plug in t=5 for each component. You can't just add them because they are perpendicular. So what is the x component of acceleration and what is the y component of acceleration at t=5? And by the way, the acceleration has units of m/sec^2.

 

[XwakeriderX]

Awesome(by the way go red soxs!) anyways haha so i have a 180 m/s^2 for the y-axis and 360 m/s^2 for the x-axis! now do i do R=sqrt(Rx^2 + Ry^2) to find the magnitude and then tan(theta)=Ry/Rx for the direction or is this completely wrong? By the way thankyou so much your making a huge difference

 

[PhanthomJay]

Awesome(by the way go red soxs!) anyways haha so i have a 180 m/s^2 for the y-axis and 360 m/s^2 for the x-axis! now do i do R=sqrt(Rx^2 + Ry^2) to find the magnitude and then tan(theta)=Ry/Rx for the direction or is this completely wrong? By the way thankyou so much your making a huge difference

Yes, that is exactly right! But now once you get that result for the acceleration, you then need to find the net force causing that acceleration. The net force is always in the direction of the acceleration. What famous law would you use to calculate the Force, once you know the mass and acceleration? (Note: I figure if the Sox win their remaining 16 games and the Yanks or Rays lose half of their remaining games , we're in!).

 

[XwakeriderX]

Newtons second law is net force= mass x acceleration... so i can use that to find the net force for x and y
Fx=(10)(360)=3600
Fy=(10)(180)=1800
hmmm now I am getting kinda lost

 

[PhanthomJay]

Newtons second law is net force= mass x acceleration... so i can use that to find the net force for x and y
Fx=(10)(360)=3600
Fy=(10)(180)=1800
hmmm now I am getting kinda lost

Where'd you get the 10..the mass is given as 15 kg. Once you make that correction, you will end up with correct value of the x and y components of the net force (in units of Newtons), and then to get the resultant net force and direction, you can do that square root and tan theta thing you did for the acceleration. OR, you can calculate the net acceleration and direction as per your prior post, and then apply Newton 2 directly to that result to get the force and direction, which must be in the same direction as the acceleration. Either way is OK.

 

[XwakeriderX]

oh wow haha i have no idea where i got the 10 i wrote the given wrong on my paper.
so now i have
Fx=5400
Fy=2700
SO!
Magnitude of force=sqrt((5400)^2 + (2700)^2)=6037.38 N (ill do sig figs later!)

Direction of force=Tan^-1(2700/5400)=26.56

 

[PhanthomJay]

oh wow haha i have no idea where i got the 10 i wrote the given wrong on my paper.
so now i have
Fx=5400
Fy=2700
SO!
Magnitude of force=sqrt((5400)^2 + (2700)^2)=6037.38 N (ill do sig figs later!)

Direction of force=Tan^-1(2700/5400)=26.56 degrees[/color]

Looks good, nice effort!

 

[XwakeriderX]

Seriously that was awesome. Thankyou so much! I am going to work on a bunch of other problems! THANKYOU THANKYOU!

 

[PhanthomJay]

Seriously that was awesome. Thankyou so much! I am going to work on a bunch of other problems! THANKYOU THANKYOU!

It was my pleasure. Wecome to PF! There are experts on this Forum in every area of science and math you can imagine, who are more than willing to assist as long as you show your relevant equations and an attempt at a solution. Come on back now, you hear?