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, we can analyze the forces acting on each block separately.
Firstly, let's consider the case where the force F is applied to M2. In this case, the force F will cause M2 to accelerate while M1 remains stationary. Since the table is frictionless, there is no horizontal friction force acting on either block.
The force F applied to M2 can be expressed as the mass M2 multiplied by its acceleration a2:
F = M2 * a2
Next, let's use Newton's second law to relate the acceleration of M2 to the contact force between the two blocks. Assuming the contact force as F_c, the net force acting on M2 is the contact force F_c minus the force F:
M2 * a2 = F_c - F
Since M1 remains stationary, its acceleration a1 will be zero. The contact force F_c acting on M1, due to M2, will be equal in magnitude but opposite in direction to the contact force acting on M2.
So, the equation for M1 becomes:
M1 * a1 = -F_c
Since a1 is zero, F_c will also be zero. Therefore, there is no contact force between the two blocks when an external force is applied only to M2.
Now let's consider the case where an equal and oppositely directed force is applied to M1. In this case, both M1 and M2 will accelerate together in the same direction as the applied force. Using the same reasoning as before, we can write the equation for M2:
M2 * a2 = F - F_c
And the equation for M1:
M1 * a1 = F_c
The two blocks will have the same acceleration a, so we can set a2 = a1 = a. Substituting this into the equations, we get:
M2 * a = F - F_c
M1 * a = F_c
Adding these two equations together, we can solve for F_c:
M1 * a + M2 * a = F + F - F_c
(F_c cancels out)
(a * M1 + a * M2) = 2F
a * (M1 + M2) = 2F
a = (2F) / (M1 + M2)
Now that we have the acceleration, we can substitute it back into one of the equations to find F_c:
F_c = M1 * a
By plugging in the given values:
a = (2 * 4.85 N) / (1.06 kg + 3.80 kg)
a ≈ 1.19 m/s^2
F_c = 1.06 kg * 1.19 m/s^2
F_c ≈ 1.26 N
Therefore, the size of the contact force between the two blocks when an equal but oppositely directed force is applied to M1 is approximately 1.26 N.