Find Magnitude & Direction for acceleration

Click For Summary

Discussion Overview

The discussion revolves around determining the magnitude and direction of acceleration for block A in relation to block B, given specific acceleration values and components. The context includes mathematical reasoning and problem-solving related to a physics homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Block B has an acceleration of 4 m/s², and the relative acceleration of block A with respect to B is also 4 m/s².
  • Some participants propose using the x and y components of acceleration, specifically -4cos(20) and -4sin(20), which yield values of -3.76 and -1.37, respectively.
  • There is a challenge regarding the application of the formula a_A = a_B + a_A/B, with some participants indicating that the substitution of values into this formula was incorrect.
  • One participant suggests expressing the acceleration vector as a combination of its components, leading to a proposed form of a_A = (4*i + (-3.76*j)) + (0*i + (-1.37*j)).
  • Another participant calculates the magnitude as |a| = sqrt((0.24^2) + (-1.37^2)) = 1.39 m/s² and the direction using inverse tangent, resulting in -80.06 degrees, although they express uncertainty about the correctness of this result.
  • One participant confirms that the calculated values align with their expectations, while another suggests sketching a triangle to visualize the vector relation, indicating that it will be isosceles.

Areas of Agreement / Disagreement

Participants express differing views on the correct application of the acceleration formula and the subsequent calculations. The discussion includes both confirmations of calculations and requests for clarification, indicating that no consensus has been reached on the method or results.

Contextual Notes

There are unresolved issues regarding the correct application of the acceleration formula and the interpretation of the components, which may depend on specific assumptions or definitions not fully articulated in the discussion.

ladkee
Messages
8
Reaction score
0

Homework Statement


Block B has acceleration of 4 m/s2... Relative acceleration of block A w/ respect to B is 4 m/s2. Find magnitude & direction of accel for A?

Homework Equations


a_A = a_B + a_A/B

x_A = x_B + x_A/B

y_A = y_B + y_A/B

The Attempt at a Solution



x & y components:
-4cos(20) = -3.76
-4sin(20) = -1.37

-3.76 = 4 + x_A/B

x_A/B = -7.76

y_A/B = -1.37

|a| = sqrt( -7.76^2 + -1.37^2 ) = 7.88 m/s2

direction: tan -1 * (-1.37/-7.88) = 10.01 degrees
 

Attachments

  • accel2.png
    accel2.png
    12.7 KB · Views: 1,603
Last edited by a moderator:
Physics news on Phys.org
ladkee said:
aa|b :
x & y components:
-4cos(20) = -3.76 ✔
-4sin(20) = -1.37 ✔

aa = ab + aa|b
x components:
-3.76 = 4 + x_A/B ✗
Hi ladkee. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Check above.
 
Last edited by a moderator:
what is the next step to find the magnitude of acceleration for a_A, are you saying use (-3.76, -1.37) as the x&y components?
 
I just added an extra line to my post. Can you fix your line I've marked with a cross?
 
NascentOxygen said:
I just added an extra line to my post. Can you fix your line I've marked with a cross?
 
NascentOxygen said:
Hi ladkee. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Check above.
I'm not sure what you mean by that? ...so are you saying for the x & y components I need to use are (-3.76, -1.37). How do you find the magnitude & direction for acceleration for this particular problem? please help, I need a clear response.
 
Last edited by a moderator:
Those are the x and y components of the smaller block's relative acceleration, yes. You worked them out.

I pointed out that you incorrectly applied the formula (which I highlighted by writing it in red). Can you apply that formula correctly to determine the x component of a's acceleration? So I'm saying there is something wrong in how you substituted values into that formula aa = ab + ...
 
NascentOxygen said:
Those are the x and y components of the smaller block's relative acceleration, yes. You worked them out.

I pointed out that you incorrectly applied the formula (which I highlighted by writing it in red). Can you apply that formula correctly to determine the x component of a's acceleration? So I'm saying there is something wrong in how you substituted values into that formula aa = ab + ...
so should it be something like ... a_A = (4*i + (- 3.76*j) ) + ( 0*i + (-1.37*j) ) = 0.24*i - 1.37*j
 
ladkee said:
so should it be something like ... a_A = (4*i + (- 3.76*j) ) + ( 0*i + (-1.37*j) ) = 0.24*i - 1.37*j
You can do it that way if you wish. Next, you're asked to express as a magnitude and direction ...
 
  • #10
NascentOxygen said:
You can do it that way if you wish. Next, you're asked to express as a magnitude and direction ...
Okay, for magnitude I did: |a| = sqrt ( (0.24^2) + (-1.37^2) = 1.39 m/s2

and direction I did inverse tangent of (y/x) which is ( -1.37 / 0.24 ) = - 80.06 deg ...which I'm not sure is right
 
  • #11
Those look to be around what I was expecting.
 
  • Like
Likes   Reactions: ladkee
  • #12
You should be able to go back to your first post in this thread, and without doing any further calculations, sketch the triangle showing this vector relation: a_A = a_B + a_A/B

Hint: The first line in your first post tells you that the triangle will be isoscles.
 
  • #13
I got the right answers, 1.39 & - 80.06, thanks for all ur help, highly appreciated
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K