Hi there. just a quick question that I want to make sure of before I try getting through all of this.
Is the Vb at this instant supposed to be zero? If yes or no, either way, is there any particular duo of points you suggest looking at first.
Thanks,
Ben
Subscribe to:
Post Comments (Atom)
15 comments:
You should NOT assume that v_B = 0.
For velocity, I recommend relating points B and C on link AC with the rigid body kinematics equation. You know the magnitude and direction of v_C and the direction of v_B. This vector equation will give you two scalar equations with v_B and omega_AC as unknowns.
(Alternately, you could find the IC for link AC. The geometry is simple enough that you can find the angular velocity of AC directly from this IC.)
Then repeat the process for acceleration relating the accelerations of B and C. a_C you know in both magnitude and direction; a_B you know the direction. Solve the two scalar equations for the unknowns of a_B and alpha_AC.
Finally, relate the acceleration of A with that of C through the acceleration equation.
Is the acceleration of C = 0? Or can we only assume that the horizontal acceleration of C = 0 and there is still a vertical component to the acceleration of C? I am getting the acceleration of A = 6.11i - 6.11j ft/s^2 and i don't know how to get rid of the horizontal component.
I am also still a bit confused on how to determine the direction of alpha.
--to the last two comments by jonbrueck--
We know that C travels on a straight line path with constant speed. Therefore, both the tangential and normal components of acceleration of C are zero.
Alpha is just one of the unknowns in the problem. You will solve for it (both magnitude and sign) from the two scalar equations coming from acceleration. My suggestion is to always write alpha_vector = alpha*k (effectively assuming that the alpha_vector is in the positive k direction). If you calculate a positive value for alpha, then this assumption is correct (positive k, or CCW). A negative value means that the assumption was wrong, and in fact, the angular acceleration is in the negative k direction (CW).
It is hard to tell how you ended up with an x-component for aA. I would be happy to look over your work if you send a scan by email.
Does this help?
When describing the radius from C to A, there is a vertical and horizontal component, right? Now that i found alpha to be in the negative k direction, the accelereration of a = 0. The alpha cross r is cancelling out with the omega squared times r.
a_A = a_C + alpha x r_A/C - omega^2*r_A/C
Since a_C = 0, this reduces to:
a_A =alpha x r_A/C - omega^2*r_A/C
I can see how some terms might cancel on the RHS. However, I would be surprised if there was a complete cancellation. That would imply that alpha x r_A/C = omega^2*r_A/C. This is NOT possible since these two vectors are perpendicular to each other -- perpendicular vectors cannot cancel (draw a picture of the vector sum of two perpendicular vectors to see this).
I would be happy to look over your solution if you send me a scan.
I can't get my scanner to work and I simply don't have any more time for this problem, so I'll just accept my failureeee.
I have the equation a_A = alpha x (-0.471i + 0.471j) - 1.08. I have broken up the equation into its i and j components, but i don't know if the -1.08, which is omega^2*r, is i or j. Any help?
I am having trouble relating the accelerations of A and B... I have written the equation a_B = a_C + (alpha_AC x r_B/C) - (omega_AC^2 * r_B/C) but this leaves me with two unknowns and one equation... What is the other equation I should use??
--to tommy worosyzlo--
It looks like you are using the equation:
a_A = a_C + omega x r_A/C - omega^2*r_A/C
where a_C = 0. In this equation, r_A/C is the VECTOR which extends from point C to point A:
r_A/C = -8*cos(45°)*i + 8*sin(45°)*j
To get to your question: you are treating "r" as a scalar (I think) -- this needs to be the VECTOR given above.
Let me know if this does not help.
--to andrew brown--
In your equation:
a_B = a_C + (alpha_AC x r_B/C) - (omega_AC^2 * r_B/C)
you are given that a_C = 0 (constant speed along a straight line path). Also you know that a_B has only a "j" component.
This vector has two scalar components -- two equations with two unknowns (a_B and alpha_AC).
Does this help?
Well I saw that you wrote that above but... I dont see how that helps me get around the fact that I have one equation and two unknowns??
--to andrew brown--
Looking at the acceleration equation:
a_B = a_C + (alpha_AC x r_B/C) - (omega_AC^2 * r_B/C)
we see that all of the terms are VECTORS: a_B, a_C, alpha_AC x r_B/C and omega_AC^2 * r_B/C.
Each vector can have up to two components: i and j. When you balance the coefficients of i and j on both sides of the equation, you will get two scalar equations.
You have two unknowns in this equation: a_B and alpha_AC. Solve these two equations for the two unknowns.
Let me know if this helps or not. This is a fundamental question -- let's get it resolved.
After finishing this problem, my answer came out to be positive and in the i direction but is the same magnitude. This is different than what is in the book. Is the book correct? If so, what could I be doing wrong? I have checked my work multiple times.
Yes, the answer in the book is correct.
I would be happy to look at your solution if you are still interested in seeing what went wrong in your analysis. You can send me a scan, or see me after class on Friday.
Post a Comment