| Problem statement Solution videohttps://youtu.be/xVRtYNh2LhM |
DISCUSSION THREAD
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DISCUSSION and HINTS
Link OA is pinned to ground at O. Slider B is constrained to move along a straight guide. Link AB, having negligible mass, connects point A on OA with slider B. With OA initially rotating CCW, and with a force F acting to the right on B, we want to know the initial acceleration of slider B.
Recall the following four-step plan outline in the lecture book and discussed in lecture:
Step 1: FBDs
Draw individual free body diagrams (FBDs) of the OA and slider B. Note that since link AB has negligible mass and that it has forces acting at only two points, AB is a two-force member. Knowing that, what are the directions of the reaction forces on OA and on B due to link AB?
Step 2: Kinetics (Newton/Euler)
You need to use an Euler (moment) equation for link OA, and a Newton equation for slider B.
Step 3: Kinematics
At the instant shown, the velocity of A is in the same direction as the velocity of B. What does this say about the instant center of AB, and therefore about the angular velocity of AB?
You will need two acceleration equations: one for link OA, and one for link AB:
aA = aO + αOA x rA/O - ωOA2 rA/O
aA = aB + αAB x rA/B - ωAB2 rA/B
Please note that ωOA is NOT known to be constant. Do not assume so. Combining the above kinematics equations will provide you with the relationships relating αOA, αAB and aB.
Step 4: Solve
From your equations in Steps 2 and 3, solve for the acceleration of the slider B.
With the instant center, if the two velocities are in the same direction, that means the perpendiculars are parallel, resulting in no instant center. This implies that the motion of AB is purely translational at this instant with zero angular velocity.
How can we find the reactions of AB at A and B? It being a two force member doesn’t make the reactions equal and along AB does it since it is moving?
Link AB has negligible mass. Although the link is accelerating, it is still a two-force member since the mass is modeled as being zero. (If you do not see this, draw an FBD of AB, and use the Newton/Euler equations...you will see that any components of reactions on AB that are perpendicular to AB are zero.)
As the link OA is homogeneous, does it mean the moment of inertia at point O is also 1/12 m(L/2)^2, or do we have to calculate it again using (IG+m(L/4)^2)?
You have to calculate it using the parallel axis theorem since it's not about the center of mass. Since you're adding an extra term, the moment of inertia about point O is larger than about the center of mass, which makes sense--it's easier spin a yardstick in the middle then it is to wave it around like a sword.
The moment of inertia about the end of a bar is 1/3m(L/2)^2 which equals 1/12(mL^2). This includes the parallel axis theorem md^2 "correction factor". You do not need to separately add that. If you take the equation for the bar about its center, which is 1/12m(L/2)^2 and add the m(L/4)^2 correction factor you will get the same value as you would if you just used the moment of inertia about its end.
Should gravity be taken into account for this problem?
Since this problem is in a horizontal plane we can just ignore gravity as it is essentially going into the page.
I see that now, thank you
Would the friction force be pointing to the right on the top and bottom of the box? I am confused because the box can slide both ways.
The surface on which the block slides is smooth. No friction.
How can we find the angular velocity of arm AB? Or is it zero because arm AB is a two-force member? If that is true, would alpha AB also be 0?
You need to use the rigid body kinematics equations for both velocity and acceleration in order to solve the problem. (You can use an IC analysis to replace the usage of the velocity kinematics equation.)
If Force AB cancels out because of angle 90, is kinematics necessary? The only unknown ends up being aB.
Kinematics is necessary for this problem, force AB ends up going away after solving for a_B, I think that is what you are asking. There are some rigid body kinematics equations given for this problem that will get you to that point.
we should draw two separate free body diagrams for this question right?
Correct. I drew a free-body diagram for link OA alone and a free-body diagram for slider B alone. I also included another diagram with the direction of the acceleration for slider B to help me set up the Newton force equations.