Chapter 4 Homework

Homework H4.N – Sp26

10 Comments

 

Problem statement
Solution video

DISCUSSION THREAD

Discussion and hints:

Recall the following four-step plan outline in the lecture book and discussed in lecture:

Step 1: FBD
Draw a free body diagram of the system made up of A+B.

Step 2: Kinetics (linear impulse/momentum and work/energy)
From your FBD above, what is the external force acting on the system of A+B in the horizontal direction? What does this say about the linear momentum of this system in that direction? Also, are there any non-conservative forces acting on the system of A+B? What does this say about the mechanical energy of the system?

Step 3: Kinematics
At position 2, B is moving only in the horizontal direction. There is no vertical component of velocity of B at position 2.

Step 4: Solve
Solve for the speeds of A and B from the above equations.

Any questions?

10 thoughts on “Homework H4.N – Sp26”

  1. To find the work done on Block A, should I make another FBD with A and B separate? Or use the FBD of A + B together and analyze it from there?

    1. It is recommended above in the Discussion that you work with a system made up of A and B together. If you follow that recommendation, then you need to work with that system throughout.

      So, for the W/E equation, include both A and B. This means that the KE will include both bodies, and the PE will include both bodies. For that system, there is no non-conservative work being done.

      And, for the LIM equation, include both A and B. For momentum, include both bodies. With A and B together, there are no external forces acting in the horizontal direction, leaving the impulse in that direction equal to zero.

      Does this help?

      1. Since we are analyzing the system A+B together, if I choose the datum (vg = 0) at the center O going horizontally, would the potential energy, V1 and V2, both be zero as the entire system has no displacement relative to the datum, or should I account for the vertical displacement of particle B when writing the work-energy equation? Thank you for the clarification.

        1. If you choose the gravitational datum line at O for the system made up of A and B together, then:

          * V_1 = 0, since the centers of mass of A and B are both on the datum line

          * V_2 = -mgR since the center of mass of A is at the datum line and B is a distance of R BELOW the datum

    1. The answer is: You ALWAYS need to define a datum line when you write down the potential due to the gravitational force. Otherwise, you cannot write down the potential.

  3. I got my V_A in position 2 as a positive value, however, the block is moving left so this answer does not make sense to me. But it seems as there is no way I get a negative value from the WE equation because the V_A will be squared no matter what so should I just stick a negative sign on my answer or maybe mention its ___ m/s in the negative I_hat direction?

    1. Take a look at the animation above. There are two instances when B is at the bottom of the slot:
      * One has B moving to the right and A moving to the left.
      * The other has B moving to the left and B moving to the right.

      Note that when you take the square root of a number, there are two roots: one positive and one negative. These two roots account to the two instances described above for the velocity of A.

      Does this help?

        1. Yes, if you choose the negative root, you are selecting the answer for A moving to the left. And, choosing the positive root gives you A moving to the right. Both are solutions to your equations and represent the motion at two different times.

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