Two blocks are in contact on a frictionless table. A horizontal force F is applied to M2, as shown. If M1 = 1.06 kg, M2 = 3.80 kg, and F = 4.85 N, find the size of the contact force between the two blocks.
If instead an equal but oppositely directed force is applied to M1 rather than M2, find the size of the contact force between the two blocks.
To find the size of the contact force between the two blocks when force F is applied to M2, we can use the equation of Newton's second law: F = ma.
Let's denote the contact force between the two blocks as FC.
For M1:
F - FC = M1 * a1
For M2:
FC = M2 * a2
Since the blocks are in contact and experiencing the same acceleration (a1 = a2 = a), we can set up a system of equations:
F - FC = M1 * a ---------(1)
FC = M2 * a ------------- (2)
Substituting equation (2) into equation (1), we get:
F - M2 * a = M1 * a
Simplifying the equation:
F = (M1 + M2) * a
Now, we can solve for the acceleration, a:
a = F / (M1 + M2)
Substituting the given values:
a = 4.85 N / (1.06 kg + 3.80 kg)
a ≈ 0.898 m/s²
Finally, we can substitute the calculated acceleration into equation (1) to find the contact force, FC:
FC = M2 * a
FC = 3.80 kg * 0.898 m/s²
FC ≈ 3.41 N
Therefore, the size of the contact force between the two blocks when force F is applied to M2 is approximately 3.41 N.
Now, let's consider the scenario where an equal but oppositely directed force is applied to M1 instead of M2.
Using the same equations, we have:
For M1:
FC - F = M1 * a1
For M2:
FC = M2 * a2
Following the same steps as before, we can substitute equation (2) into equation (1) to solve for the contact force, FC:
FC - F = M1 * a
M2 * a = F
Simplifying the equation:
FC = M1 * a + F
Substituting the given values:
FC = 1.06 kg * 0.898 m/s² + 4.85 N
FC ≈ 6.31 N
Therefore, the size of the contact force between the two blocks when an equal but oppositely directed force is applied to M1 instead of M2 is approximately 6.31 N.
To find the size of the contact force between the two blocks, we need to first understand the behavior of the system.
In the given scenario, the blocks are in contact on a frictionless table. When a force is applied to one of the blocks, it creates an acceleration in that block and subsequently affects the other block due to their contact.
Let's consider the first scenario where a horizontal force (F) is applied to M2. In this case, M2 experiences an acceleration, and the contact force between the two blocks is the force that M1 exerts on M2.
We can use Newton's second law of motion to find the acceleration of M2:
F = m2 * a2
Rearranging the equation, we can solve for the acceleration (a2) of M2:
a2 = F / m2
Now that we have the acceleration of M2, we can find the contact force (Fc) between the two blocks. The contact force will be equal in magnitude but opposite in direction to the force M1 exerts on M2.
Fc = -m1 * a2
Substituting the values given, we have:
Fc = -(1.06 kg) * (F / 3.80 kg)
Now, let's move on to the second scenario, where an equal but oppositely directed force is applied to M1. In this case, M1 experiences an acceleration, and the contact force between the two blocks is the force that M2 exerts on M1. Following the same steps as before, we can find the contact force between the blocks.
The acceleration of M1 can be found using Newton's second law:
F = m1 * a1
Rearranging the equation, we can solve for the acceleration (a1) of M1:
a1 = F / m1
The contact force between the blocks is then equal in magnitude but opposite in direction to the force M2 exerts on M1.
Fc = -m2 * a1
Substituting the values given, we have:
Fc = -(3.80 kg) * (F / 1.06 kg)
Thus, by following the steps explained above, you can calculate the size of the contact force between the two blocks for both scenarios.