# Find Magnitude & Direction for acceleration

## 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

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## Answers and Replies

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NascentOxygen
Staff Emeritus
Science Advisor
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 [Broken]

Check above.

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

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

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

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.

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NascentOxygen
Staff Emeritus
Science Advisor
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 + ...

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

NascentOxygen
Staff Emeritus
Science Advisor
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 ....

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
Staff Emeritus
Science Advisor
Those look to be around what I was expecting.

• ladkee
NascentOxygen
Staff Emeritus
Science Advisor
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.

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