Here you can add discussion posts as well as add comments to existing posts. Links are also provided to solution videos and other pages of the course website for your easy access.
Suggestions Place observer on the ladder. Also, align the x-axis with the ladder. With these choices: * omega = theta_dot*k * the relative velocity is b_dot*i * the relative acceleration is b_dot_dot*i
Points A and B are not on the same rigid body (distance between the two points is increasing with time). Therefore, you will not able to use the rigid body kinematics equations between those two points.
So, it is not a question of using the new equations for the sake of using them; the old ones just don't work.
On some problems in this section where there are a number of links involved, you are able to (and encouraged to) use the rigid body equations on a rigid link. You have one of those due on Monday, and we did one in class on Friday.
when we are finding the acceleration of A with respect to the truck, do we assume the acceleration of the observer is zero, since it is moving with the truck?
For part a) you are asked to find the acceleration of A relative to the truck. Since B is on the truck, B has the same acceleration as the truck. Therefore for part a) you are asked to find:
a_A/B = a_A - a_B = (a_A/B)_rel + alpha x r_B/A +2*omega x (v_A/B)_rel + omega x (omega x r_A/B)
You need the observer in order to find (a_A/B)_rel and (v_A/B)_rel for this equation. The omega term in the equation is the angular velocity of the observer.
For part b) you are asked to find the acceleration of A with respect to ground. Use your result above for a_A/B to get:
The terms that go into the answer for part a) are:
a_A/B = (a_A/B)_rel + alpha x r_B/A +2*omega x (v_A/B)_rel + omega x (omega x r_A/B)
For these terms: * (a_A/B)_rel is in the negative x-component * alpha x r_B/A is zero (constant omega) * 2*omega x (v_A/B)_rel is in the positive y-direction (omega is positive k and (v_A/B)_rel is positive x. Therefore, k x i = +j) * omega x (omega x r_A/B) is in the negative x-direction
Based on this, it looks like a) should have a negative x-component and a positive y-component.
I might have misunderstood your question. I read the question as to if the answer in part a) would have both negative x- and y- components. Based on my arguments that I presented, I would conclude that the answer for a) will have a negative x-component and a positive y-component.
To get the answer for part b), you just add on the acceleration of B. a_B has a negative x-component and a positive y-component. Therefore, adding on a_B will not change the signs of the components.
Getting back to your omega question. omega appears in two terms; the Coriolis term and the omega x (omega x r_A/B) term. The Coriolis term is postive y and the other term is negative x, as described in the earlier post.
I may have done something wrong but I didn't use the velocity of the truck to solve the problem. Do you not need to know that? Or, did I do something wrong?
I must have done something wrong but I set the x axis horizontal with the page and not going in the direction of the ladder so we aren't really on the same page with the equations and such.
15 comments:
would it be better to use the rigid body equations here or are we supposed to be using moving reference frame equations now?
Points A and B are not on the same rigid body (distance between the two points is increasing with time). Therefore, you will not able to use the rigid body kinematics equations between those two points.
So, it is not a question of using the new equations for the sake of using them; the old ones just don't work.
On some problems in this section where there are a number of links involved, you are able to (and encouraged to) use the rigid body equations on a rigid link. You have one of those due on Monday, and we did one in class on Friday.
if we place the observer on the ladder itself, won't that give us the wrong answer since we are finding the acceleration with respect to the truck?
when we are finding the acceleration of A with respect to the truck, do we assume the acceleration of the observer is zero, since it is moving with the truck?
I got a negative answer for both the i and j components of the relative acceleration. is that possible since theta_dot is counterclockwise?
is the v_b= 35 mi/hr and not zero, or is b a pin point?
--to jonbrueck--
Good question.
For part a) you are asked to find the acceleration of A relative to the truck. Since B is on the truck, B has the same acceleration as the truck. Therefore for part a) you are asked to find:
a_A/B = a_A - a_B = (a_A/B)_rel + alpha x r_B/A +2*omega x (v_A/B)_rel + omega x (omega x r_A/B)
You need the observer in order to find (a_A/B)_rel and (v_A/B)_rel for this equation. The omega term in the equation is the angular velocity of the observer.
For part b) you are asked to find the acceleration of A with respect to ground. Use your result above for a_A/B to get:
a_A = a_B + a_A/B
Does this help?
--to wendy--
What you are saying is correct for part a). See the above comment that I posted for jonbrueck.
--to jonbrueck--
The terms that go into the answer for part a) are:
a_A/B = (a_A/B)_rel + alpha x r_B/A +2*omega x (v_A/B)_rel + omega x (omega x r_A/B)
For these terms:
* (a_A/B)_rel is in the negative x-component
* alpha x r_B/A is zero (constant omega)
* 2*omega x (v_A/B)_rel is in the positive y-direction (omega is positive k and (v_A/B)_rel is positive x. Therefore, k x i = +j)
* omega x (omega x r_A/B) is in the negative x-direction
Based on this, it looks like a) should have a negative x-component and a positive y-component.
Let me know if this does not help.
--to deborah-
Yes, v_B = 35 mph (B is a pin attached to the truck so it has the same motion as the truck).
Be sure to convert this to feet/second.
i don't really understand what you are saying about the omega term. Does that mean the acceleration vector can have both negative components?
--to jonbrueck--
I might have misunderstood your question. I read the question as to if the answer in part a) would have both negative x- and y- components. Based on my arguments that I presented, I would conclude that the answer for a) will have a negative x-component and a positive y-component.
To get the answer for part b), you just add on the acceleration of B. a_B has a negative x-component and a positive y-component. Therefore, adding on a_B will not change the signs of the components.
Getting back to your omega question. omega appears in two terms; the Coriolis term and the omega x (omega x r_A/B) term. The Coriolis term is postive y and the other term is negative x, as described in the earlier post.
Let me know if I missed your question.
I may have done something wrong but I didn't use the velocity of the truck to solve the problem. Do you not need to know that? Or, did I do something wrong?
The velocity of the truck is not needed for finding the acceleration of A or for finding the acceleration of A relative to the truck. You're OK.
I must have done something wrong but I set the x axis horizontal with the page and not going in the direction of the ladder so we aren't really on the same page with the equations and such.
Post a Comment