A hand pushes three identical bricks as shown. The bricks are moving to the left and speeding up. System A consists of two bricks stacked together. System B consists of a single brick. System C consists of all three bricks. There is friction between the bricks and the table.
Mech Newton’s second and third laws HW-42 3. A hand pushes three identical bricks as shown. The bricks are moving to the left and speeding up. System A consists of two bricks stacked together. System B consists of a single brick. System C consists of all three bricks. There is friction between the bricks and the table. a. In the spaces provided at right, draw and label separate free-body diagrams for systems A and B. PAD Ñ₂ BH Free-body diagram for system A NA 2mg b. The vector representing the acceleration of system A is shown at right. Draw the acceleration vectors for systems B and C using the same scale. Explain. Zm c. The vector representing the net force on system A is shown at right. Draw the net force vectors for systems B and C using the same scale. Explain. Ñ AB F Ñâ‚� BA JAT A Free-body diagram for system B NB Explain how you knew to draw the force vectors as you did. mg d. The vector representing the frictional force on system A is shown below. Draw the remaining force vectors using the same scale. B Acceleration of A RAB FB Acceleration of B Acceleration of C Net force on A Net force on B Net force on C fBT
The correct answer and explanation is:
Here’s how to approach the problem step by step, addressing each part:
Part (a): Free-body diagrams for systems A and B
System A (Two stacked bricks):
- Normal force (NAN_A): Acts vertically upward, balancing the weight of two bricks.
- Weight (2mg2mg): Acts vertically downward; this is the combined weight of the two bricks.
- Friction (fAf_A): Acts horizontally to the right, opposing the motion.
- Applied force (FhandF_{hand}): Acts horizontally to the left, accelerating the system.
System B (Single brick):
- Normal force (NBN_B): Acts vertically upward, balancing the weight of one brick.
- Weight (mgmg): Acts vertically downward; this is the weight of the single brick.
- Friction (fBf_B): Acts horizontally to the right, opposing the motion.
- Interaction force (FABF_{AB}): A horizontal force exerted by system A on system B. This force accelerates system B.
Part (b): Acceleration vectors
- All systems have the same acceleration because they move together as a single group.
- Therefore:
- The acceleration vector for system B should have the same direction and magnitude as the one given for system A.
- The acceleration vector for system C (all three bricks combined) is also identical to that of systems A and B.
Part (c): Net force vectors
- The net force is directly proportional to the total mass of each system, given by Fnet=maF_{net} = ma.
- For system A (2m2m), the net force vector has been drawn.
- For system B (mm), the net force vector should be half the length of the vector for system A, as its mass is half.
- For system C (3m3m), the net force vector should be 1.5 times the length of the vector for system A, as its mass is 1.5 times larger.
Part (d): Frictional force and other forces
- System A: Frictional force (fA=2μmgf_A = 2\mu mg), since two bricks exert double the normal force of one brick.
- System B: Frictional force (fB=μmgf_B = \mu mg).
- System C: Frictional force (fC=3μmgf_C = 3\mu mg), as all three bricks contribute to the normal force.
The applied force must overcome the total friction force (Fapplied>fCF_{applied} > f_C) and provide the net force required to accelerate the system.
Explanation Summary
To analyze the problem, we used Newton’s Second Law (Fnet=maF_{net} = ma) to relate the forces and accelerations of the systems. The free-body diagrams for systems A and B include vertical forces (NN and mgmg) and horizontal forces (friction and applied/interaction forces). Acceleration is identical across systems because the bricks move together as one group, implying they share the same rate of change of velocity.
Net force vectors are scaled according to each system’s total mass. For system A (2m2m), the net force is proportional to 2ma2ma; for system B (mm), it’s proportional to mama; and for system C (3m3m), it’s proportional to 3ma3ma. This scaling ensures consistency with the total force required to accelerate each system.
Frictional forces depend on the normal force, which is proportional to the weight of the bricks (f=μNf = \mu N). System A experiences double the friction of system B because it has twice the weight. System C, which combines all three bricks, has the largest frictional force. The applied force must overcome this total frictional force and still provide the net force required to accelerate the bricks.
Understanding the relationships between mass, force, and acceleration provides a complete solution to the problem. Each system’s motion and force interactions align with Newton’s Laws, showing how individual components contribute to the overall dynamics.