Three blocks on a frictionless horizontal surface are in contact with each other, as shown below. A force F is applied to block A (mass mA).
(a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system (in terms of mA, mB, mC), (c) the net force on each block, and (d) the force of contact that each block exerts on its neighbor. (e) If mA = mB = mC =12.0kg and F=96.0N, give numerical answers to (b),(c), and(d). Do your answers make sense intuitively?
Diagram:
>|mA||mB||mC|
(a) Free-body diagrams:
Block A (mA):
- The applied force F is acting towards the right.
- There is a force of contact between block A and block B, directed towards the right.
- There is also a force of contact between block A and the surface, directed towards the left (opposite to the applied force).
- Since the surface is frictionless, there are no forces related to friction.
Block B (mB):
- There is a force of contact between block B and block A, directed towards the left.
- There is a force of contact between block B and block C, directed towards the right.
- There is no force acting vertically on block B.
- Since the surface is frictionless, there are no forces related to friction.
Block C (mC):
- There is a force of contact between block C and block B, directed towards the left.
- There is a force of contact between block C and the surface, directed towards the right.
- There is no force acting vertically on block C.
- Since the surface is frictionless, there are no forces related to friction.
(b) Acceleration of the system:
To find the acceleration of the system, we can use Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration (F = ma).
Since the three blocks are in contact with each other and have no external forces acting on them other than the applied force F, the net force experienced by the system is simply equal to the applied force F.
The total mass of the system is given by the sum of the masses of the three blocks (m_total = mA + mB + mC). Therefore, the acceleration of the system can be calculated as:
a = F / m_total
(c) Net force on each block:
Since the blocks are in contact with each other, the net force on each block is equal to the force of contact exerted by its neighbor.
For block A:
Net force = Force of contact with B (towards the right) = FB_A
For block B:
Net force = Force of contact with A (towards the left) + Force of contact with C (towards the right) = FA_B - FC_B
For block C:
Net force = Force of contact with B (towards the left) = FB_C
(d) Force of contact between each block:
Based on Newton's third law of motion, the force of contact between two objects is equal in magnitude but opposite in direction.
For block A:
Force of contact with B = FB_A (towards the right)
For block B:
Force of contact with A = FA_B (towards the left)
Force of contact with C = FB_C (towards the right)
For block C:
Force of contact with B = FC_B (towards the left)
(e) Numerical answers:
If mA = mB = mC = 12.0 kg and F = 96.0 N, we can substitute these values into the calculations from parts (b), (c), and (d) to find the numerical answers.
(b) Acceleration of the system:
m_total = mA + mB + mC = 12.0 kg + 12.0 kg + 12.0 kg = 36.0 kg
a = F / m_total = 96.0 N / 36.0 kg = 2.67 m/s^2
(c) Net force on each block:
For block A: Net force = FB_A = F = 96.0 N
For block B: Net force = FA_B - FC_B = 0 N - 0 N = 0 N
For block C: Net force = FB_C = 0 N
(d) Force of contact between each block:
For block A: Force of contact with B = FB_A = 96.0 N
For block B: Force of contact with A = FA_B = 0 N, Force of contact with C = FB_C = 0 N
For block C: Force of contact with B = FC_B = 0 N
Intuitively, the answers make sense as the acceleration of the system is proportional to the applied force and inversely proportional to the total mass of the system. Since the mass of block A is the same as the masses of blocks B and C, it makes sense that the net force on block A is equal to the applied force and that there are no net forces on blocks B and C. The force of contact between the blocks also appears consistent.
(a) Free-body diagrams for each block:
Block A (mA):
- Force of contact with block B
- Force of contact with block C
- Applied force F
Block B (mB):
- Force of contact with block A
- Force of contact with block C
Block C (mC):
- Force of contact with block B
- Force of contact with block A
(b) The acceleration of the system:
To determine the acceleration, we need to apply Newton's second law individually to each block. The sum of the external forces on each block equals mass times acceleration.
For block A (mA):
Sum of external forces = Force of contact with block B + Applied force F
mA * acceleration = Force of contact with block B + F
For block B (mB):
Sum of external forces = Force of contact with block A + Force of contact with block C
mB * acceleration = Force of contact with block A + Force of contact with block C
For block C (mC):
Sum of external forces = Force of contact with block B + Force of contact with block A
mC * acceleration = Force of contact with block B + Force of contact with block A
Since the blocks are in contact with each other and there is no friction, the magnitude of the force of contact between each pair of blocks is the same. Let's denote it as F_contact.
From the above equations, and since the force of contact between each block pair is the same, we can rewrite them as:
mB * acceleration = F_contact + F_contact
mA * acceleration = F_contact + F
mC * acceleration = F_contact + F_contact
Simplifying these equations, we get:
mB * acceleration = 2 * F_contact
mA * acceleration = F_contact + F
mC * acceleration = 2 * F_contact
(c) The net force on each block:
Since the only external force acting on the system is the applied force F, the net force on each block is equal to the applied force.
For block A:
Net force = F
For block B:
Net force = F
For block C:
Net force = F
(d) The force of contact that each block exerts on its neighbor:
Since the blocks are in contact with each other and there is no friction, the magnitude of the force of contact between each pair of blocks is the same. Let's denote it as F_contact.
For block A:
Force of contact with block B = F_contact
Force of contact with block C = F_contact
For block B:
Force of contact with block A = F_contact
Force of contact with block C = F_contact
For block C:
Force of contact with block A = F_contact
Force of contact with block B = F_contact
(e) If mA = mB = mC = 12.0 kg and F = 96.0 N:
Substituting the given values into the equations, we can calculate the numerical values.
For (b) the acceleration of the system:
Using the equation: mB * acceleration = 2 * F_contact
12.0 kg * acceleration = 2 * F_contact
acceleration = 2 * F_contact / 12.0 kg
For (c) the net force on each block:
The net force on each block is equal to the applied force, which is 96.0 N.
For (d) the force of contact that each block exerts on its neighbor:
The force of contact between each pair of blocks is equal to half the applied force, which is 48.0 N.
Intuitively, the acceleration of the system should be inversely proportional to the mass of the blocks. As the mass increases, the acceleration should decrease. Additionally, the net force on each block and the force of contact between each pair of blocks should be equal in magnitude, as there is no friction or external forces acting on the blocks.