Homework H2.F - Sp23

Problem statement
Solution video
https://youtu.be/lHCWDbNV9a4

DISCUSSION THREAD

Ask your questions here. And, consider answering questions of your colleagues here. Either way, you can learn.


DISCUSSION and HINTS

In this problem, end A of the bar is constrained to move along a straight horizontal path with a constant speed of vA, whereas end B is constrained to move along a straight, angled path. As you can see in the animation below of the motion of the bar, the speed of B is NOT a constant (the acceleration of B is non-zero, and is, in fact, increasing as B moves along its path).

In your solution, it is recommended that you use the rigid body kinematics equations relating the motion of ends A and B:

vB = vA + ω x rB/A
aB = aA + α x rB/A - ω2rB/A

For these equations, you know: i) the magnitude and direction for the velocity of A; ii) that the acceleration of A is zero (constant speed along a straight path); and, iii) the direction for the velocity and acceleration of B. These two vector equations produce four scalar equations that can be solved for four scalar unknowns: vB, aB, ω and α.

 

17 thoughts on “Homework H2.F - Sp23”

  1. For this problem, I set up the velocity equations for B with respect to A then with respect to O, which is the point on the ground where the angle is 45. I did this because we needed another equation since there's too many unknowns for one equation. However, in the end you technically get a value for the angular velocity of the rigid body that O is on (the line that B slides on), but isn't it supposed to be zero since it's stationary? I know this answer doesn't ask to solve for it but you'll need it later on when solving for acceleration in the same method...unless this method is incorrect, in which case I'm confused on how to solve for the angular terms.

    1. I dont think were solving for the angular terms, they're just being used as a variable that you solve for to use in the equations for the normal velocity and acceleration

    1. I drew a vertical line down from B so that it's perpendicular to the ground. From there, you'll have to use trig, the angles, and the L of the bar to solve for the x and y components of B

    2. You can set up two triangles using the given angles phi and theta and length L and use trig and hypotenuse theorem to solve for the i and j components. It is helpful to note that since phi = 45º, two of the sides are the same length.

    3. I used the phi angle for B's position vector. We know from the hints that the velocity and acceleration are going down across the wire. One component of the velocity would be Vb * cos (phi) for example.

  2. Given the angle phi as equal to 45 degrees, the i and j components for Vb should be equal to each other, same for Ab. that gives you four equations to find four unknowns.

  3. I'm pretty sure you cannot relate B to a fixed point other than A, because B is moving and they are not a fixed distance apart. The equations are only valid for points on the same rigid body.

  4. When I first read through the problem I had no idea how the angle phi was going to be used. Like Micheal pointed out above in this case the magnitudes of the i and j components of Vb are equal, but I think it is important to note that if we were given a different angle such as 60 degrees we would still simply be able to write Vb in component form using trig and keeping our equations equal to our unknowns in that way.

  5. My initial thought was to use phi for the position vector of B, but I realized theta must be used because the position vector is obviously with respect to A. Phi is used later on when balancing the i and j components of Vb. Sin(phi) and cos(phi) are equivalent, so the i and j components can be set equal, which will lead us to finding one of our unknowns.

    1. That would be true if v_B were constant. It is given that v_A is constant; however, that does not say that v_B = constant. Therefore, a_B is probably not zero.

  6. I'am still confused in what format the answers should be in. Since the question doesn't say to solve in terms of vectors does that mean to solve for the exact value or it doesn't matter?

Leave a Reply