Find Magnitude & Direction for acceleration

[ladkee]

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

 

[NascentOxygen]

a_a|b :
x & y components:
-4cos(20) = -3.76 ✔
-4sin(20) = -1.37 ✔

a_a = a_b + a_a|b
x components:
-3.76 = 4 + x_A/B ✗

Hi ladkee. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Check above.

 

[ladkee]

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?

 

[NascentOxygen]

I just added an extra line to my post. Can you fix your line I've marked with a cross?

 

[ladkee]

I just added an extra line to my post. Can you fix your line I've marked with a cross?

 

[ladkee]

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.

 

[NascentOxygen]

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 a_a = a_b + ...

 

[ladkee]

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 a_a = a_b + ...

so should it be something like ... a_A = (4*i + (- 3.76*j) ) + ( 0*i + (-1.37*j) ) = 0.24*i - 1.37*j

 

[NascentOxygen]

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 ...

 

[ladkee]

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

 

[NascentOxygen]

Those look to be around what I was expecting.

 

[NascentOxygen]

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.

 

[ladkee]

I got the right answers, 1.39 & - 80.06, thanks for all ur help, highly appreciated